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If three coins are tossed, represent the sample space and the event of getting at least two heads, then find the number of elements in them.
Let \(S\) be the sample space and \(E\) be the event of occurrence of atleast two heads and let \(H\) denote the occurrence of head and \(T\) denote the occurrence of tail, when one coin is tossed.
Then,
\(
\begin{aligned}
& S=\{H, T\} \times\{H, T\} \times\{H, T\} \\
& S=\{(H, H, H),(H, H, T),(H, T, H),(T, H, H), \\
& \quad(T, T, H),(T, H, T),(H, T, T),(T, T, T)\}
\end{aligned}
\)
and \(E=\{(H, H, H),(H, H, T),(H, T, H),(T, H, H)\}\)
Also, \(n(S)=8\) and \(n(E)=4\)
One ticket is drawn at random from a bag containing 24 tickets numbered 1 to 24 . Represent the sample space and the event of drawing a ticket containing number which is a prime. Also, find the number of elements in them.
Let \(S\) be the sample space and \(E\) be the event of occurrence a prime number.
Then,
\(
\begin{aligned}
& S=\{1,2,3,4,5, \ldots, 24\} \\
& E=\{2,3,5,7,11,13,17,19,23\}
\end{aligned}
\)
and
Also, \(\quad n(S)=24\) and \(n(E)=9\)
Two dice are thrown simultaneously. What is the probability obtaining a total score less than 11?
Let \(S\) be the sample space and \(E\) be the event of obtaining a total less than 11 .
Then, \(S=\{1,2,3,4,5,6\} \times\{1,2,3,4,5,6\} \Rightarrow n(S)=6 \times 6=36\)
Let \(E^{\prime}\) be the event of obtaining a total score greater than or equal to 11 .
Also,
\(
E^{\prime}=\{(5,6),(6,5),(6,6)\} ; \quad \therefore n\left(E^{\prime}\right)=3
\)
Then, probability of obtaining a total score greater than or equal to 11 ,
\(
\begin{aligned}
& P\left(E^{\prime}\right)=\frac{n\left(E^{\prime}\right)}{n(S)}=\frac{3}{36}=\frac{1}{12} \\
\therefore \quad & P(E)=1-P\left(E^{\prime}\right)=1-\frac{1}{12}=\frac{11}{12}
\end{aligned}
\)
Hence, required probability is \(\frac{11}{12}\).
If a leap-leap year is selected at random, then what is the chance it will contain 53 Sunday?
1 year \(=365\) days
A leap year has 366 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks \(=52 \times 7=364\) days
\(366-364=2\) days
In a leap year there will be 52 Sundays and 2 days will be left.
These 2 days can be:
1. Sunday, Monday
2. Monday, Tuesday
3. Tuesday, Wednesday
4. Wednesday, Thursday
5. Thursday, Friday
6. Friday, Saturday
7. Saturday, Sunday
Of these total 7 outcomes, the favourable outcomes are 2.
Hence the probability of getting 53 days \(=\frac{2}{7}\)
From a pack of 52 playing cards, three cards are drawn at random. Find the probability of drawing a King, a Queen and a Knave.
Let \(S\) be the sample space and \(E\) be the event that out of the three cards drawn one is a King, one is a Queen and one is a Knave.
\(\therefore n(S)=\) Total number of selecting 3 cards out of 52 cards
\(
={ }^{52} C_3
\)
and \(n(E)=\) Number of selecting 3 cards out of one is King, one is Queen and one is Knave \(={ }^4 C_1 \cdot{ }^4 C_1 \cdot{ }^4 C_1=64\)
\(\therefore\) Required probability, \(P(E)=\frac{n(E)}{n(S)}=\frac{64}{{ }^{52} C_3}=\frac{\frac{64}{52 \cdot 51 \cdot 50}}{1 \cdot 2 \cdot 3}=\frac{16}{5525}\)
A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that
(i) all the three balls are white.
(ii) all the three balls are red.
(iii) one ball is red and two balls are white.
Let \(S\) be the sample space, \(E_1\) be the event of getting 3 white balls, \(E_2\) be the event of getting 3 red balls and \(E_3\) be the event of getting one red ball and two white balls.
\(\therefore n(S)=\) Number of ways of selecting 3 balls out of \(13(8+5)={ }^{13} C_3=\frac{13 \cdot 12 \cdot 11}{1 \cdot 2 \cdot 3}=286\)
(i) \(n\left(E_1\right)=\) Number of ways of selecting 3 white balls out of 5
\(
={ }^5 C_3={ }^5 C_2=\frac{5 \cdot 4}{1 \cdot 2}=10
\)
\(
\therefore P \text { (getting } 3 \text { white balls })=\frac{n\left(E_1\right)}{n(S)}=\frac{10}{286}=\frac{5}{143}
\)
(ii) \(
\begin{aligned}
n\left(E_2\right) & =\text { Number of ways of selecting } 3 \text { red balls out of } 8 \\
& ={ }^8 C_3=\frac{8 \cdot 7 \cdot 6}{1 \cdot 2 \cdot 3}=56
\end{aligned}
\)
\(
\therefore P(\text { getting } 3 \text { red balls })=\frac{n\left(E_2\right)}{n(S)}=\frac{56}{286}=\frac{28}{143}
\)
(iii) \(n\left(E_3\right)=\) Number of ways of selecting 1 red ball out of 8 and 2 black balls out of \(5={ }^8 C_1 \cdot{ }^5 C_2=8 \cdot 10=80\)
\(\therefore P\) (getting 1 red and 2 black balls)
\(
=\frac{n\left(E_3\right)}{n(S)}=\frac{80}{286}=\frac{40}{143}
\)
For a post, three persons \(A, B\) and \(C\) appear in the interiew. The probability of \(A\) being selected is twice that of \(B\) and the probability of \(B\) being selected is thrice that of \(C\). What are the individual probabilities of \(A, B\) and \(C\) being selected?
Let \(E_1, E_2\) and \(E_3\) be the events of selection of \(A, B\) and \(C\) respectively.
Let \(P\left(E_3\right)=x\).
Then, \(P\left(E_2\right)=3 P\left(E_3\right)=3 x\) and \(P\left(E_1\right)=2 P\left(E_2\right)=6 x\) Since, \(E_1, E_2\) and \(E_3\) are mutually exclusive and exhaustive events.
\(
\begin{aligned}
& \therefore & P\left(E_1 \cup E_2 \cup E_3\right) & =P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1 \\
& \therefore & P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right) & =1 \\
& \Rightarrow & 6 x+3 x+x & =1 \\
& \therefore & x & =\frac{1}{10}
\end{aligned}
\)
Hence,
\(
\begin{aligned}
& P\left(E_1\right)=6 x=\frac{6}{10}=\frac{3}{5} \\
& P\left(E_2\right)=3 x=\frac{3}{10} \text { and } P\left(E_3\right)=x=\frac{1}{10}
\end{aligned}
\)
If \(A\) and \(B\) are independent events, the probability that both \(A\) and \(B\) occur is \(\frac{1}{8}\) and the probability that none of them occurs is \(\frac{3}{8}\). Find the probability of the occurrence of \(A\).
We have,
\(
P(A \cap B)=\frac{1}{8} \Rightarrow P(A) P(B)=\frac{1}{8} \dots(i)
\)
[ \(\because A\) and \(B\) are independent]
\(
\begin{aligned}
\text { and } & P(\bar{A} \cap \bar{B})=\frac{3}{8} \Rightarrow P(\bar{A}) P(\bar{B}) =\frac{3}{8} \\
\Rightarrow & (1-P(A))(1-P(B)) =\frac{3}{8} \\
\Rightarrow & 1-P(A)-P(B)+\frac{1}{8} =\frac{3}{8} \text { [from Eq. (i)] } \\
\Rightarrow & P(A)+P(B) =\frac{3}{4} \dots(ii)
\end{aligned}
\)
The quadratic equation whose roots are \(P(A)\) and \(P(B)\) is
\(x^2-[P(A)+P(B)] x+P(A) \cdot P(B)=0\) \(\Rightarrow \quad x^2-\frac{3}{4} x+\frac{1}{8}=0\) [from Eqs. (i) and (ii)]
or \(8 x^2-6 x+1=0\) or \(x=\frac{1}{2}, \frac{1}{4}\)
Hence, \(\quad P(A)=\frac{1}{2}\) or \(\frac{1}{4}\)
\(A\) and \(B\) are two candidates seeking admission in IIT. The probability that \(A\) is selected is 0.5 and the probability that both \(A\) and \(B\) are selected is atmost 0.3 . Is it possible that the probability of \(B\) getting selected is 0.9?
\(
\text { Let } E_1 \text { and } E_2 \text { are the events of } A \text { and } B \text { selected, respectively. }
\)
\(
\begin{aligned}
& \text { Given, } P\left(E_1 \cap E_2\right) \leq 0.3 \text { and } P\left(E_1\right)=0.5 \\
& \text { Since, } P\left(E_1 \cup E_2\right)=P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\
& \because \quad P\left(E_1 \cup E_2\right) \leq 1 \\
& \therefore \quad P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \leq 1
\end{aligned}
\)
\(
\begin{array}{lr}
\Rightarrow & P\left(E_1\right)+P\left(E_2\right) \leq 1+P\left(E_1 \cap E_2\right) \\
\Rightarrow & 0.5+P\left(E_2\right) \leq 1+0.3 \Rightarrow P\left(E_2\right) \leq 0.8
\end{array}
\)
\(
\text { Hence, } \quad P\left(E_2\right) \neq 0.9
\)
Let \(A, B\) and \(C\) be three events. If the probability of occurring exactly one event out of \(A\) and \(B\) is \(1-a\), out of \(B\) and \(C\) is \(1-2 a\), out of \(C\) and \(A\) is \(1-a\) and that of occurring three events simultaneously is \(a^2\), then what is the probability that atleast one out of \(A, B\) and \(C\) will occur is greater than ____.
Given
\(
P(A)+P(B)-2 P(A \cap B)=1-a \dots(i)
\)
\(
P(B)+P(C)-2 P(B \cap C)=1-2 a \dots(ii)
\)
\(
\text { and } \quad P(C)+P(A)-2 P(C \cap A)=1-a \dots(iii)
\)
\(
\therefore \quad P(A \cap B \cap C)=a^2 \dots(iv)
\)
\(
\begin{aligned}
\therefore P(A \cup B \cup C)= & P(A)+P(B)+P(C)-P(A \cap B) \\
& -P(B \cap C)-P(C \cap A)+P(A \cap B \cap C)
\end{aligned}
\)
\(
\begin{aligned}
=\frac{1}{2}\{P(A)+ & P(B)-2 P(A \cap B)+P(B)+P(C)-2 P(B \cap C) \\
& +P(C)+P(A)-2 P(C \cap A)\}+P(A \cap B \cap C)
\end{aligned}
\)
\(
=\frac{1}{2}\{1-a+1-2 a+1-a\}+a^2 \text { [from Eqs. (i), (ii), (iii) and (iv)] }
\)
\(
=\frac{3}{2}-2 a+a^2=(a-1)^2+\frac{1}{2}>\frac{1}{2} \quad[\because a \neq 1]
\)
Let \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be three events such that \(\mathrm{P}(\mathrm{A})=0.3 \mathrm{P}(\mathrm{B})=0.4, \mathrm{P}(\mathrm{C})=0.8, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\) \(0.08, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=0.28, \mathrm{P}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=0.09\). If \(\mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C}) \geq 0.75\), then \(\mathrm{P}(\mathrm{B} \cap \mathrm{C})\) lies in what interval?
Let \(\mathrm{P}(\mathrm{B} \cap \mathrm{C})=x\)
Since,
\(
\begin{aligned}
P(A \cup B \cup C)= & P(A)+P(B)+P(C)-P(A \cap B) \\
& -P(B \cap C)-P(C \cap A)+P(A \cup B \cup C)
\end{aligned}
\)
\(
\therefore \quad=0.3+0.4+0.8-0.08-x-0.28+0.09=1.23-x
\)
But given that, \(P(A \cup B \cup C) \geq 0.75\) and \(P(A \cup B \cup C) \leq 1\)
\(
\begin{aligned}
& \therefore \quad 0.75 \leq 1.23-x \leq 1 \Rightarrow-0.75 \geq-1.23+x \geq-1 \\
& \text { or } \quad 1.23-0.75 \geq x \geq 1.23-1 \text { or } 0.23 \leq x \leq 0.48
\end{aligned}
\)
Two dice are thrown. Find the probability that the sum of the numbers coming up on them is 9 , if it is known that the number 5 always occurs on the first dice.
Let \(S\) be the sample space
\(
\begin{aligned}
& \therefore \quad S=\{1,2,3,4,5,6\} \times\{1,2,3,4,5,6\} \\
& \therefore \quad n(S)=36 \\
&
\end{aligned}
\)
and let \(E_1 \equiv\) The event that the sum of the numbers coming up is 9 .
and \(\quad E_2 \equiv\) The event of occurrence of 5 on the first dice.
\(
\begin{aligned}
& \therefore \quad E_1 \equiv\{(3,6),(6,3),(4,5),(5,4)\} \\
& \therefore \quad n\left(E_1\right)=4 \\
&
\end{aligned}
\)
\(
\begin{aligned}
& \text { and } \quad E_2=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\} \\
& \therefore n\left(E_2\right)=6 \\
& E_1 \cap E_2=\{(5,4)\} \\
& \therefore \quad n\left(E_1 \cap E_2\right)=1 \\
& \text { Now, } \quad P\left(E_1 \cap E_2\right)=\frac{n\left(E_1 \cap E_2\right)}{n(S)}=\frac{1}{36} \\
& \text { and } \quad P\left(E_2\right)=\frac{n\left(E_2\right)}{n(S)}=\frac{6}{36}=\frac{1}{6} \\
&
\end{aligned}
\)
\(\therefore\) Required probability,
\(
\begin{aligned}
& P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6} \\
\end{aligned}
\)
\(
\text { Alternate:} \quad P\left(\frac{E_1}{E_2}\right)=\frac{n\left(E_1 \cap E_2\right)}{n\left(E_2\right)}=\frac{1}{6}
\)
In a class, \(30 \%\) students fail in English; \(20 \%\) students fail in Hindi and \(10 \%\) students fail in English and Hindi both. A student is chosen at random, then what is the probability that he will fail in English, if he has failed in Hindi?
Let \(S\) be the sample space.
If \(n(S)=100\), then
\(E_1 \equiv\) The event that the student chosen fail in English
\(
\therefore n\left(E_1\right)=30
\)
and \(E_2 \equiv\) The event that the student chosen fail in Hindi
\(
\therefore n\left(E_2\right)=20 \text { and } n\left(E_1 \cap E_2\right)=10
\)
\(
\begin{aligned}
\therefore \quad P\left(E_2\right) & =\frac{n\left(E_2\right)}{n(S)} \\
& =\frac{20}{100}=\frac{1}{5}
\end{aligned}
\)
and \(P\left(E_1 \cap E_2\right)=\frac{n\left(E_1 \cap E_2\right)}{n(S)}=\frac{10}{100}=\frac{1}{10}\)
\(\therefore\) Required probability, \(P\left(\frac{E_1}{E_2}\right)=\frac{P\left(E_1 \cap E_2\right)}{P\left(E_2\right)}=\frac{\frac{1}{10}}{\frac{1}{5}}=\frac{1}{2}\)
Alternate:
\(
\begin{aligned}
P\left(\frac{E_1}{E_2}\right) & =\frac{n\left(E_1 \cap E_2\right)}{n\left(E_2\right)} \\
& =\frac{10}{20}=\frac{1}{2}
\end{aligned}
\)
The probability that certain electronic component fails, when first used is 0.10 . If it does not fail immediately, then the probability that it lasts for one year is 0.99 . What is the probability that a new component will last for one year?
Given probability of electronic component fails, when first used \(=0.10\)
i.e., \(\quad P(\underline{F})=0.10\)
\(
\therefore \quad P(\bar{F})=1-P(F)=0.90
\)
and let \(P(Y)=\) Probability of new component to last for one year
\(
\because \quad P(F)+P(\bar{F})=1
\)
Obviously, the two events are mutually exclusive and exhaustive
\(
\begin{aligned}
\therefore \quad P\left(\frac{Y}{F}\right) & =0 \text { and } P\left(\frac{Y}{\bar{F}}\right)=0.99 \\
\therefore \quad P(Y) & =P(F) \cdot P\left(\frac{Y}{F}\right)+P(\bar{F}) \cdot P\left(\frac{Y}{\bar{F}}\right) \\
& =0.10 \times 0+0.90 \times 0.99 \\
& =0+(0.9)(0.99)=0.891
\end{aligned}
\)
Three groups \(A, B\) and \(C\) are contesting for positions on the Board of Directors of a company. The probabilities of their winning are \(0.5,0.3\) and 0.2 , respectively. If the group \(A\) wins, then the probability of introducing a new product is 0.7 and the corresponding probabilities for groups \(B\) and \(C\) are 0.6 and 0.5 , respectively. Find the probability that the new product will be introduced.
Given, \(P(A)=0.5, P(B)=0.3\) and \(P(C)=0.2\)
\(\therefore \quad P(A)+P(B)+P(C)=1\)
Then, events \(A, B, C\) are exhaustive.
If \(P(E)=\) Probability of introducing a new product, then as given
\(
\begin{aligned}
P\left(\frac{E}{A}\right) & =0.7, P\left(\frac{E}{B}\right)=0.6 \text { and } P\left(\frac{E}{C}\right)=0.5 \\
P(E) & =P(A) \cdot P\left(\frac{E}{A}\right)+P(B) \cdot P\left(\frac{E}{B}\right)+P(C) \cdot P\left(\frac{E}{C}\right) \\
& =0.5 \times 0.7+0.3 \times 0.6+0.2 \times 0.5 \\
& =0.35+0.18+0.10=0.63
\end{aligned}
\)
An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white, it is not replace into urn, otherwise it is replaced along with another ball of the same colour. The process is repeated, find the probability that the third ball drawn is black.
For the first two draw, the balls taken out may be Let \(E_1=\) White and White;
\(\quad E_2=\) White and Black \(E_3=\) Black and White \(; \quad E_4=\) Black and Black
\(
\begin{aligned}
\therefore \quad & P\left(E_1\right)=P(W) \cdot P\left(\frac{W}{W}\right)=\frac{2}{4} \cdot \frac{1}{3}=\frac{1}{6} \\
& P\left(E_2\right)=P(W) \cdot P\left(\frac{B}{W}\right)=\frac{2}{4} \cdot \frac{2}{3}=\frac{1}{3} \\
& P\left(E_3\right)=P(B) \cdot P\left(\frac{W}{B}\right)=\frac{2}{4} \cdot \frac{2}{5}=\frac{1}{5}
\end{aligned}
\)
\(
\text { and } P\left(E_4\right)=P(B) \cdot P\left(\frac{B}{B}\right)=\frac{2}{4} \cdot \frac{3}{5}=\frac{3}{10}
\)
\(
\begin{aligned}
\therefore P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)+P\left(E_4\right) & =\frac{1}{6}+\frac{1}{3}+\frac{1}{5}+\frac{3}{10} \\
& =\frac{10+20+12+18}{60}=1
\end{aligned}
\)
Then, events \(E_1, E_2, E_3\) and \(E_4\) are exhaustive. Obviously, these events are mutually exclusive, then
\(
\begin{aligned}
& P\left(\frac{B}{E_1}\right)=\frac{2}{2}=1 ; P\left(\frac{B}{E_2}\right)=\frac{3}{4} \\
& P\left(\frac{B}{E_3}\right)=\frac{3}{4} \text { and } P\left(\frac{B}{E_4}\right)=\frac{4}{6}=\frac{2}{3}
\end{aligned}
\)
\(\therefore\) Required probability,
\(
P(B)=P\left(E_1\right) \cdot P\left(\frac{B}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{B}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{B}{E_3}\right)+P\left(E_4\right) \cdot P\left(\frac{B}{E_4}\right)
\)
\(
\begin{aligned}
& =\frac{1}{6} \times 1+\frac{1}{3} \times \frac{3}{4}+\frac{1}{5} \times \frac{3}{4}+\frac{3}{10} \times \frac{2}{3} \\
& =\frac{1}{6}+\frac{1}{4}+\frac{3}{20}+\frac{1}{5} \\
& =\frac{10+15+9+12}{60}=\frac{46}{60}=\frac{23}{30}
\end{aligned}
\)
A bag A contains 2 white and 3 red balls and a bag \(B\) contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and it is found to be red. Then, find the probability that it was drawn from the bag \(B\).
Let \(E_1 \equiv\) The event of ball being drawn from bag \(A\).
\(E_2 \equiv\) The event of ball being drawn from bag \(B\).
and \(E \equiv\) The event of ball being red.
Since, both the bag are equally likely to be selected, therefore
\(
P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2} \text { and } P\left(\frac{E}{E_1}\right)=\frac{3}{5} \text { and } P\left(\frac{E}{E_2}\right)=\frac{5}{9}
\)
\(\therefore\) Required probability,
\(
\begin{aligned}
P\left(\frac{E_2}{E}\right) & =\frac{P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)} \\
& =\frac{\frac{1}{2} \times \frac{5}{9}}{\frac{1}{2} \times \frac{3}{5}+\frac{1}{2} \times \frac{5}{9}}=\frac{\frac{9}{9}}{\frac{3}{5}+\frac{5}{9}}=\frac{25}{52}
\end{aligned}
\)
A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
Then, \(P(E)=\frac{1}{6}\)
\(\therefore \quad P(\bar{E})=1-P(E)=1-\frac{1}{6}=\frac{5}{6}\)
\(\therefore \quad P\left(\frac{E_1}{E}\right)=P(\) man speaking the truth \()=\frac{3}{4}\)
and \(P\left(\frac{E_1}{\bar{E}}\right)=P(\) man not speaking the truth \()=1-\frac{3}{4}=\frac{1}{4}\)
Clearly, \(\left(\frac{E}{E_1}\right)\) is the event that it is actually a six, when it is known that the man reports a six.
\(
\begin{aligned}
P\left(\frac{E}{E_1}\right) & =\frac{P(E) \cdot P\left(\frac{E_1}{E}\right)}{P(E) \cdot P\left(\frac{E_1}{E}\right)+P(\bar{E}) \cdot P\left(\frac{E_1}{E}\right)} \\
& =\frac{\frac{1}{6} \times \frac{3}{4}}{\frac{1}{6} \times \frac{3}{4}+\frac{5}{6} \times \frac{1}{4}}=\frac{3}{8}
\end{aligned}
\)
In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is \(\frac{1}{3}\) and the probability that he copies the answer is \(\frac{1}{6}\). The probability that his answer is correct given that he copied it is \(\frac{1}{8}\). Find the probability that he knew the answer to the question given that he correctly answered it.
Lest \(E_1\) be the event that the answer is guessed, \(E_2\) be the event that the answer is copied, \(E_3\) be the event that the examinee knows the answer and \(E\) be the event that the examinee answers correctly.
Given, \(P\left(E_1\right)=\frac{1}{3}, P\left(E_2\right)=\frac{1}{6}\)
Assume that events \(E_1, E_2\) and \(E_3\) are exhaustive
\(
\begin{array}{cc}
\therefore & P\left(E_1\right)+P\left(E_2\right)+P\left(E_3\right)=1 \\
\therefore & P\left(E_3\right)=1-P\left(E_1\right)-P\left(E_2\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2}
\end{array}
\)
Now, \(P\left(\frac{E}{E_1}\right)\)
\(\equiv\) Probability of getting correct answer by guessing
\(
=\frac{1}{4}
\)
[since 4 alternatives]
\(P\left(\frac{E}{E_2}\right) \equiv\) Probability of answering correctly by copying \(=\frac{1}{8}\)
and \(P\left(\frac{E}{E_3}\right) \equiv\) Probability of answering correctly by
knowing \(=1\)
Clearly, \(\left(\frac{E_3}{E}\right)\) is the event he knew the answer to the question, given that he correctly answered it.
\(
\begin{aligned}
\therefore P\left(\frac{E_3}{E}\right) & =\frac{P\left(E_3\right) \cdot P\left(\frac{E}{E_3}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_3}\right)} \\
& =\frac{\frac{1}{2} \times 1}{\frac{1}{3} \times \frac{1}{4}+\frac{1}{6} \times \frac{1}{8}+\frac{1}{2} \times 1}=\frac{24}{29}
\end{aligned}
\)
\(A\) and \(B\) are two independent witnesses (i.e., there is no collusion between them) in a case. The probability that \(A\) will speak the truth is \(x\) and the probability that \(B\) will speak the truth is \(y\). \(A\) and \(B\) agree in a certain statements. Find the probability that the statements is true
Let \(E_1\) be the event that both \(A\) and \(B\) speak the truth, \(E_2\) be the event that both \(A\) and \(B\) tell a lie and \(E\) be the event that \(A\) and \(B\) agree in a certain statements.
And also, let \(C\) be the event that \(A\) speak the truth and \(D\) be the event that \(B\) speaks the truth.
\(
\therefore \quad E_1=C \cap D
\)
\([\because C\) and \(D\) are independent events]
\(
\text { and } \quad E_2=\bar{C} \cap \bar{D}
\)
\(
\begin{aligned}
& \text { then, } \quad P\left(E_1\right)=(C \cap D)=P(C) \cdot P(D)=x y \\
& \text { and } \quad P\left(E_2\right)=P(\bar{C} \cap \bar{D})=P(\bar{C}) P(\bar{D}) \\
& =\{1-P(C)\}\{1-P(D)\}=(1-x)(1-y) \\
& =1-x-y-x y \\
&
\end{aligned}
\)
Now, \(P\left(\frac{E}{E_1}\right) \equiv\) Probability that \(A\) and \(B\) will agree, when both of them speak the truth \(=1\)
and \(P\left(\frac{E}{E_2}\right)=\) Probability that \(A\) and \(B\) will agree, when both of them tell a lie \(=1\)
Clearly, \(\left(\frac{E_1}{E}\right)\) be the event that the statement is true
\(
\begin{aligned}
\therefore P\left(\frac{E_1}{E}\right) & =\frac{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)}{P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right)} \\
& =\frac{x y \cdot 1}{x y \cdot 1+(1-x-y+x y) \cdot 1}=\frac{x y}{1-x-y+2 x y}
\end{aligned}
\)
If on an average, out of 10 ships, one is drowned, then what is the probability that out of 5 ships, atleast 4 reach safely?
Let \(p\) be the probability that a ship reaches safely. \(\therefore p=\frac{9}{10}\)
\(\therefore q=\) Probability that a ship is drowned \(=1-p=1-\frac{9}{10}\)
\(
\therefore \quad q=\frac{1}{10}
\)
Let \(X\) be the random variable, showing the number of ships reaching safely.
Then, \(P\) (atleast 4 reaching safely) \(=P(X=4\) or \(X=5)\)
\(
=P(X=4)+P(X=5)
\)
\(
={ }^5 C_4\left(\frac{9}{10}\right)^4\left(\frac{1}{10}\right)^{5-4}+{ }^5 C_5\left(\frac{9}{10}\right)^5\left(\frac{1}{10}\right)^{5-5}
\)
\(
=\frac{5 \times 9^4}{10^5}+\frac{9^5}{10^5}=\frac{9^4 \times 14}{10^5}
\)
Numbers are selected at random one at a time, from the numbers \(00,01,02, \ldots, 99\) with replacement. An event \(E\) occurs, if and only if the product of the two digits of a selected number is 18 . If four numbers are selected, then find the probability that E occurs atleast 3 times.
Out of the numbers \(00,01,02, \ldots, 99\), those numbers the product of whose digits is 18 are \(29,36,63,92\) i.e., only 4 .
\(
p=P(E)=\frac{4}{100}=\frac{1}{25}, q=P(\bar{E})=1-\frac{1}{25}=\frac{24}{25}
\)
Let \(X\) be the random variable, showing the number of times E occurs in 4 selections.
Then, \(P(E\) occurs atleast 3 times \()=P(X=3\) or \(X=4)\)
\(
\begin{aligned}
& =P(X=3)+P(X=4)={ }^4 C_3 p^3 q^1+{ }^4 C_4 p^4 q^0 \\
& =4 p^3 q+p^4=4 \times\left(\frac{1}{25}\right)^3 \times \frac{24}{25}+\left(\frac{1}{25}\right)^4 \\
& =\frac{97}{390625}
\end{aligned}
\)
A man takes a step forward with probability 0.4 and backward with probability 0.6 . Then, find the probability that at the end of eleven steps he is one step away from the starting point.
Since, the man is one step away from starting point mean that either
(i) man has taken 6 steps forward and 5 steps backward.
(ii) man has taken 5 steps forward and 6 steps backward. Taking, movement 1 step forward as success and 1 step backward as failure.
\(
\begin{aligned}
& \therefore \quad p=\text { Probability of success }=0.4 \\
& \text { and } q=\text { Probability of failure }=0.6
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \text { Required probability }=P(X=6 \text { or } X=5) \\
& \quad=P(X=6)+P(X=5)={ }^{11} C_6 p^6 q^5+{ }^{11} C_5 p^5 q^6 \\
& \quad={ }^{11} C_5\left(p^6 q^5+p^5 q^6\right) \\
& \quad=\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}\left\{(0 \cdot 4)^6(0 \cdot 6)^5+(0 \cdot 4)^5(0 \cdot 6)^6\right\} \\
& \quad=\frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}(0 \cdot 24)^5=0 \cdot 37
\end{aligned}
\)
Hence, the required probability is 0.37.
Find the minimum number of tosses of a pair of dice, so that the probability of getting the sum of the digits on the dice equal to 7 on atleast one toss, is greater than 0.95 . (Given, \(\log _{10} 2=0 \cdot 3010\), \(\left.\log _{10} 3=0.4771\right)\)
The sample space,
\(
S=\{1,2,3,4,5,6\} \times\{1,2,3,4,5,6\}
\)
\(\therefore \quad n(S)=36\) and let \(E\) be the event getting the sum of digits on the dice equal to 7 , then
\(
\begin{aligned}
E & =\{(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)\} \\
\quad n(E) & =6 \\
p & =\text { Probability of getting the sum } 7 \\
p & =\frac{6}{36}=\frac{1}{6} \therefore q=1-p=1-\frac{1}{6}=\frac{5}{6}
\end{aligned}
\)
\(\because\) Probability of not throwing the sum 7 in first \(m\) trials \(=q^m\)
\(\therefore P\) (atleast one 7 in \(m\) throws \()=1-q^m=1-\left(\frac{5}{6}\right)^m\)
\(
\text { According to the question, } 1-\left(\frac{5}{6}\right)^m>0.95
\)
\(
\left(\frac{5}{6}\right)^m<1-0.95 \Rightarrow\left(\frac{5}{6}\right)^m<0.05
\)
\(
\left(\frac{5}{6}\right)^m<\frac{1}{20}
\)
Taking logarithm,
\(
\begin{aligned}
& \Rightarrow \quad m\left\{\log _{10} 5-\log _{10} 6\right\}<\log _{10} 1-\log _{10} 20 \\
& \Rightarrow m\left\{1-\log _{10} 2-\log _{10} 2-\log _{10} 3\right\}<0-\log _{10} 2-\log _{10} 10 \\
& \Rightarrow \quad m\left\{1-2 \log _{10} 2-\log _{10} 3\right\}<-\log _{10} 2-1 \\
& \Rightarrow \quad m\{1-0.6020-0.4771\}<-0.3010-1 \\
& \Rightarrow \quad-0.079 m<-1.3010 \\
& \Rightarrow \quad m>\frac{1.3010}{0.079}=16.44 \\
& \therefore \quad m>16.44 \\
&
\end{aligned}
\)
Hence, the least number of trials is 17 .
What is the probability distribution, when three coins are tossed.
Let \(X\) be a random variable denoting the number of heads occurred, then \(P(X=0)=\) Probability of occurrence of zero head
\(
=P(T T T)=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}
\)
\(\begin{aligned} P(X=1) & =\text { Probability of occurrence of one head } \\ & =P(H T T)+P(T H T)+P(T T H)\end{aligned}\)
\(
=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{3}{8}
\)
\(
\begin{aligned}
P(X=2) & =\text { Probability of occurrence of two heads } \\
& =P(H H T)+P(H T H)+P(T H H) \\
& =\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{3}{8}
\end{aligned}
\)
\(
\begin{aligned}
P(X=3) & =\text { Probability of occurrence of three heads } \\
& =P(H H H)=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}
\end{aligned}
\)
Thus, the probability distribution when three coins are tossed is as given below
\(
\begin{array}{c|c|c|c|c}
\hline X & 0 & 1 & 2 & 3 \\
\hline P(X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \\
\hline
\end{array}
\)
The mean and variance of a binomial variable \(X\) are 2 and 1, respectively. Find the probability that \(X\) takes values greater than 1.
Given, mean, \(n p=2 \dots(i)\)
and variance, \(n p q=1 \dots(ii)\)
On dividing Eq. (ii) by Eq. (i), we get \(q=\frac{1}{2}\) \(\therefore \quad p=1-q=\frac{1}{2}\)
\(
\therefore \quad p=1-q=\frac{1}{2}
\)
\(
\text { From Eq. (i), } n \times \frac{1}{2}=2 \therefore n=4
\)
The binomial distribution is \(\left(\frac{1}{2}+\frac{1}{2}\right)^4\)
Now,
\(
\begin{aligned}
P(X>1) & =P(X=2)+P(X=3)+P(X=4) \\
& ={ }^4 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^2+{ }^4 C_4\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^3+{ }^4 C_4\left(\frac{1}{2}\right)^4 \\
& =\frac{6+4+1}{16}=\frac{11}{16}
\end{aligned}
\)
\(
\begin{aligned}
& \text { Alternate: } P(X>1)=1-\{P(X=0)+P(X=1)\} \\
& \quad=1-\left\{{ }^4 C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^4+{ }^4 C_3\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)^3\right\}=1-\left(\frac{1+4}{16}\right)=\frac{11}{16}
\end{aligned}
\)
A random variable \(X\) has Poisson’s distribution with mean 3 . Then find the value of \(P(X>2.5)\)
\(
P(X>2.5)=1-P(X=0)-P(X=1)-P(X=2)
\)
\(
\because \quad P(X=k)=e^{-\lambda} \cdot \frac{\lambda^k}{k !}
\)
\(
\therefore \quad P(X>2.5)=1-\frac{e^{-\lambda}}{0 !}-\frac{e^{-\lambda} \cdot \lambda^1}{1 !}-\frac{e^{-\lambda} \cdot \lambda^2}{2 !}
\)
\(
\begin{aligned}
& =1-e^{-\lambda}\left(1+\lambda+\frac{\lambda^2}{2}\right) \\
& =1-e^{-3}\left(1+3+\frac{9}{2}\right) \quad(\because \lambda=n p=3)
\end{aligned}
\)
\(
=1-\frac{17}{2 e^3}
\)
A and B throw with one die for a stake of ₹ 11 which is to be won by the player who first throw 6. If \(A\) has the first throw, then what are their respective expectations?
Since, \(A\) can win the game at the 1 st, 3 rd, 5 th,…, trials. If \(p\) be the probability of success and \(q\) be the probability of fail, then
\(
p=\frac{1}{6} \text { and } q=\frac{5}{6}
\)
\(
\begin{aligned}
& P(A \text { wins at the first trial })=\frac{1}{6} \\
& P(A \text { wins at the } 3 \text { rd trials })=\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \\
& P(A \text { wins at the } 5 \text { th trials })=\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \text { and so on. }
\end{aligned}
\)
\(
\text { Therefore, } P(A \text { wins })=\frac{1}{6}+\left(\frac{5}{6}\right)^2 \frac{1}{6}+\left(\frac{5}{6}\right)^4 \cdot \frac{1}{6}+\ldots \infty
\)
\(
=\frac{\frac{1}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{6}{11}
\)
\(
\text { Similarly, } P(B \text { wins })=\frac{5}{6} \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^3 \frac{1}{6}+\left(\frac{5}{6}\right)^5 \frac{1}{6}+\ldots \infty
\)
\(
=\frac{\frac{5}{6} \cdot \frac{1}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{5}{11}
\)
Hence, expectations of \(A\) and \(B\) are \(₹ \frac{6}{11} \times 11\) and \(₹ \frac{5}{11} \times 11\), respectively. i.e. Expectations of \(A\) and \(B\) are \(₹ 6\) and \(₹ 5\), respectively.
A person throws two dice, one the common cube and the other a regular tetrahedron, the number on the lowest face being taken in the case of the tetrahedron, then find the probability that the sum of the numbers appearing on the dice is 6 .
Let \(S\) be the sample space, then
\(
\begin{aligned}
S & =\{1,2,3,4\} \times\{1,2,3,4,5,6\} \\
\therefore \quad n(S) & =24
\end{aligned}
\)
If \(E\) be the event that the sum of the numbers on dice is 6 . Then, \(n(E)=\) Coefficient of \(x^6\) in
\(
\begin{aligned}
\left(x^1+x^2+x^3+x^4\right) \times\left(x^1+\right. & \left.x^2+x^3+x^4+x^5+x^6\right) \\
& =1+1+1+1=4 \\
\therefore \text { Required probability, } P(E) & =\frac{n(E)}{n(S)}=\frac{4}{24}=\frac{1}{6}
\end{aligned}
\)
Five ordinary dice are rolled at random and the sum of the numbers shown on them is 16 . What is the probability that the numbers shown on each is any one from 2,3,4 or 5 ?
If the integers \(x_1, x_2, x_3, x_4\) and \(x_5\) are shown on the dice, then \(x_1+x_2+x_3+x_4+x_5=16\)
where, \(1 \leq x_i \leq 6\) \((i=1,2,3,4,5)\)
The number of total solutions of this equation.
\(
\begin{aligned}
& =\text { Coefficient of } x^{16} \text { in }\left(x^1+x^2+x^3+x^4+x^5+x^6\right)^5 \\
& =\text { Coefficient of } x^{16} \text { in } x^5\left(1+x+x^2+x^3+x^4+x^5\right)^5 \\
& =\text { Coefficient of } x^{11} \text { in }\left(1+x+x^2+x^3+x^4+x^5\right)^5
\end{aligned}
\)
\(
\begin{aligned}
& =\text { Coefficient of } x^{11} \text { in }\left\{\left(\frac{1-x^6}{1-x}\right)^5\right\} \\
& =\text { Coefficient of } x^{11} \text { in }\left(1-x^6\right)^5(1-x)^{-5} \\
& =\text { Coefficient of } x^{11} \text { in }
\end{aligned}
\)
\(
\left(1-5 x^6+\ldots\right)\left(1+{ }^5 C_1 x+{ }^6 C_2 x^2+\ldots\right.\left.+{ }^9 C_{C_5} x^5+\ldots+{ }^{15} C_{11} x^{11}+\ldots\right)
\)
\(
={ }^{15} C_{11}-5 \cdot{ }^9 C_5
\)
\(
={ }^{15} C_4-5 \cdot{ }^9 C_4=\frac{15 \cdot 14 \cdot 13 \cdot 12}{1 \cdot 2 \cdot 3 \cdot 4}-5 \cdot \frac{9 \cdot 8 \cdot 7 \cdot 6}{1 \cdot 2 \cdot 3 \cdot 4}=735
\)
If \(S\) be the sample space, \(\therefore \quad n(S)=735\)
Let \(E\) be the occurrence event, then
\(n(E)=\) The number of integral solutions of
\(
x_1+x_2+x_3+x_4+x_5=16 \text {, }
\)
\(
\text { where } 2 \leq x_i \leq 5 \quad(i=1,2,3,4,5)
\)
\(
\begin{aligned}
& =\text { Coefficient of } x^{16} \text { in }\left(x^2+x^3+x^4+x^5\right)^5 \\
& =\text { Coefficient of } x^{16} \text { in } x^{10}\left(1+x+x^2+x^3\right)^5 \\
& =\text { Coefficient of } x^6 \text { in }\left(1+x+x^2+x^3\right)^5 \\
& =\text { Coefficient of } x^6 \text { in }\left\{\left(\frac{1-x^4}{1-x}\right)^5\right\} \\
& =\text { Coefficient of } x^6 \text { in }\left(1-x^4\right)^5(1-x)^{-5} \\
& =\text { Coefficient of } x^6 \text { in }
\end{aligned}
\)
\(
\left(1-5 x^4+\ldots\right)\left(1+{ }^5 C_1 x+{ }^6 C_2 x^2+\ldots+{ }^{10} C_6 x^6+\ldots\right)
\)
\(
\begin{aligned}
& ={ }^{10} C_6-5 \cdot{ }^6 C_2={ }^{10} C_4-5 \cdot{ }^6 C_2 \\
& =\frac{10 \cdot 9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3 \cdot 4}-5 \cdot \frac{6 \cdot 5}{1 \cdot 2}=210-75=135
\end{aligned}
\)
\(
\therefore \text { The required probability, } P(E)=\frac{n(E)}{n(S)}=\frac{135}{735}=\frac{9}{49}
\)
The probability that in a year of 22nd century chosen at random, there will be 53 Sundays, is
(d) In the 22 nd century, there are 25 leap years viz. 2100 , \(2104,2108, \ldots, 2196\) and 75 non-leap years.
Consider the following events:
\(E_1=\) Selecting a leap year from 22 nd century
\(E_2=\) Selecting a non-leap year from 22 nd century
\(E=\) There are 53 Sundays in a year of 22 nd century
We have,
\(
P\left(E_1\right)=\frac{25}{100}, P\left(E_2\right)=\frac{75}{100}, P\left(\frac{E}{E_1}\right)=\frac{2}{7} \text { and } P\left(\frac{E}{E_2}\right)=\frac{1}{7}
\)
\(
\begin{aligned}
& \text { Required probability }=P(E)=P\left(\left(E \cap E_1\right) \cup\left(E \cap E_2\right)\right) \\
&=P\left(E \cap E_1\right)+P\left(E \cap E_2\right) \\
&=P\left(E_1\right) \cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{E}{E_2}\right) \\
&=\frac{25}{100} \times \frac{2}{7}+\frac{75}{100} \times \frac{1}{7}=\frac{5}{28}
\end{aligned}
\)
In a convex hexagon two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the hexagon, is
(a) We have,
Number of diagonals of a hexagon \(={ }^6 C_2-6=9\)
\(\therefore \quad n(s)=\) Total number of selections of two diagonals
\(
={ }^9 C_2=36
\)
and \(n(E)=\) The number of selections of two diagonals which intersect at an interior point
\(=\) The number of selections of four vertices \(={ }^6 C_4=15\)
Hence, required probability \(=\frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)
If three integers are chosen at random from the set of first 20 natural numbers, the chance that their product is a multiple of 3 , is
(d) \(n(S)=\) Total number of ways of selecting 3 integers from 20 natural numbers \(={ }^{20} C_3=1140\).
Their product is multiple of 3 means atleast one number is divisible by 3 . The number which are divisible by 3 are 3,6 , \(9,12,15\) and 16 .
\(\therefore n(E)=\) The number of ways of selecting atleast one of them multiple of 3
\(
\begin{aligned}
& \qquad={ }^6 C_1 \times{ }^{14} C_2+{ }^6 C_2 \times{ }^{14} C_1 + { }^6 C_3=776 \\
& \therefore \text { Required probability }=\frac{n(E)}{n(S)} \\
& \qquad=\frac{776}{1140}=\frac{194}{285}
\end{aligned}
\)
If three numbers are selected from the set of the first 20 natural numbers, the probability that they are in \(G P\), is
(c) \(n(S)=\) Total number of ways of selecting 3 numbers from first 20 natural numbers \(={ }^{20} C_3=1140\)
Three numbers are in GP, the favourable cases are \(1,2,4 ; 1\), 3,\(9 ; 1,4,16 ; 2,4,8 ; 2,6,18 ; 3,6,12 ; 4,8,16 ; 5,10,20 ; 4,6,9\); \(8,12,18 ; 9,12,16\)
\(\therefore n(E)=\) The number of favourable cases \(=11\)
\(\therefore\) Required probability \(=\frac{n(E)}{n(S)}=\frac{11}{1140}\)
Two numbers \(b\) and \(c\) are chosen at random with replacement from the numbers \(1,2,3,4,5,6,7,8\) and 9 . The probability that \(x^2+b x+c>0\) for all \(x \in R\), is
(b) Here, \(x^2+b x+c>0, \forall x \in R\)
\(
\begin{array}{lrl}
\therefore & D<0 \\
\Rightarrow & b^2<4 c
\end{array}
\)
\(
\begin{array}{ccll}
\hline \text { Value of } \boldsymbol{b} \text { Possible values of } \boldsymbol{c} & & \\
\hline 1 & 1<4 c & \Rightarrow c>\frac{1}{4} \Rightarrow & \{1,2,3,4,5,6,7,8,9\} \\
\hline 2 & 4<4 c & \Rightarrow c>1 \Rightarrow & \{2,3,4,5,6,7,8,9\} \\
\hline 3 & 9<4 c & \Rightarrow>\frac{9}{4} \Rightarrow & \{3,4,5,6,7,8,9\} \\
\hline 4 & 16<4 c & \Rightarrow c>4 \Rightarrow & \{5,6,7,8,9\} \\
\hline 5 & 25<4 c & \Rightarrow c>6.25 \Rightarrow & \{7,8,9\} \\
\hline 6 & 36<4 c & \Rightarrow c>9 \Rightarrow & \text { Impossible } \\
\hline 7 & \text { Impossible } \\
\hline 8 & \text { Impossible } \\
\hline 9 & \text { Impossible } \\
\hline
\end{array}
\)
\(\begin{aligned} & n(E)=\text { Number of favourable cases }=9+8+7+5+3=32 \\ & n(S)=\text { Total ways }=9 \times 9=81 \\ & \therefore \text { Required probability }=\frac{n(E)}{n(S)}=\frac{32}{81}\end{aligned}\)
Three dice are thrown. The probability of getting a sum which is a perfect square, is
(d) \(n(S)=\) Total number of ways \(=6 \times 6 \times 6=216\)
The sum of the numbers on three dice varies from 3 to 18 and among these 4,9 and 16 are perfect squares.
\(
\begin{aligned}
& \therefore n(E)=\text { Number of favourable ways } \\
& =\text { Coefficient of } x^4 \text { in }
\end{aligned}
\)
\(
\begin{aligned}
& \left(x+x^2+\ldots+x^6\right)^3+\text { Coefficient of } x^9 \text { in } \\
& \left(x+x^2+\ldots+x^6\right)^3+\text { Coefficient of } x^{16} \text { in } \\
& \left(x+x^2+\ldots+x^6\right)^3 \\
& =\text { Coefficient of } x \text { in }\left(1+x+\ldots+x^5\right)^3+\text { Coefficient of } x^6
\end{aligned}
\)
\(
\begin{aligned}
& \text { in }\left(1+x+x^2+\ldots+x^5\right)^3+\text { Coefficient of } x^{13} \\
& \text { in }\left(1+x+x^2+\ldots+x^5\right)^3 \\
& =\text { Coefficient of } x \text { in }\left(1-x^6\right)^3(1-x)^{-3}+\text { Coefficient of } x^6 \text { in } \\
& \left(1-x^6\right)^3(1-x)^{-3}+\text { Coefficient of } x^{13} \text { in }\left(1-x^6\right)^3(1-x)^{-3} \\
& =\text { Coefficient of } x \text { in }(1)\left(1+{ }^3 C_1 x+\ldots\right)+\text { Coefficient of } x^6
\end{aligned}
\)
\(
\begin{aligned}
& \text { in }\left(1-3 x^6\right)\left(1+{ }^3 C_1 x+\ldots\right)+\text { Coefficient of } x^{13} \text { in } \\
& \left(1-3 x^6+3 x^{12}+\ldots\right) ;\left(1+{ }^3 C_1 x+\ldots\right) \\
& ={ }^3 C_1+\left({ }^8 C_6-3\right)+\left({ }^{15} C_{13}-3 \times{ }^9 C_7+9\right) \\
& ={ }^3 C_1+\left({ }^8 C_2-3\right)+\left({ }^{15} C_2-3 \times{ }^9 C_2+9\right) \\
& =3+25+6 \\
& =34 \\
& \therefore \text { Required probability }=\frac{n(E)}{n(S)}=\frac{34}{216}=\frac{17}{108}
\end{aligned}
\)
A quadratic equation is chosen from the set of all quadratic equations which are unchanged by squaring their roots. The chance that the chosen equation has equal roots, is
(a) Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation. According to question,
\(
\alpha+\beta=\alpha^2+\beta^2 \text { and } \alpha \beta=\alpha^2 \beta^2 \Rightarrow \alpha \beta(\alpha \beta-1)=0
\)
\(
\Rightarrow \alpha \beta=1 \text { or } \alpha \beta=0
\)
\(
\Rightarrow \alpha=1, \beta=1 ; \alpha=\omega \beta=\omega^2 \quad \text { [cube roots and unity] }
\)
\(
\alpha=1, \beta=0 ; \alpha=0, \beta=0
\)
\(\therefore n(S)=\) Number of quadratic equations which are unchanged by squaring their roots \(=4\)
and \(n(E)=\) Number of quadratic equations have equal roots \(=2\)
\(\therefore\) Required probability \(=\frac{n(E)}{n(S)}=\frac{2}{4}=\frac{1}{2}\)
Three-digit numbers are formed using the digits 0 , \(1,2,3,4,5\) without repetition of digits. If a number is chosen at random, then the probability that the digits either increase or decrease, is
(c) \(n(S)=\) Total number of three digit numbers
\(
={ }^6 P_3-{ }^5 P_2=120-20=100
\)
\(n(E)=\) Number of numbers with digits either increase or decrease
\(=\) Number of numbers with increasing digits + Number of numbers with decreasing digits
\(
={ }^5 C_3+{ }^6 C_3=10+20=30
\)
\(\therefore\) Required probability \(=\frac{n(E)}{n(S)}=\frac{30}{100}=\frac{3}{10}\)
If \(X\) follows a binomial distribution with parameters \(n=8\) and \(p=\frac{1}{2}\), then \(p(|x-4| \leq 2)\) is equal to
(b) Here, \(p=\frac{1}{2}, n=8\)
\(\therefore \quad q=1-p=1-\frac{1}{2}=\frac{1}{2}\)
\(\therefore\) The binomial distribution is \(\left(\frac{1}{2}+\frac{1}{2}\right)^8\)
Also, \(\quad|x-4| \leq 2\)
\(\Rightarrow \quad-2 \leq x-4 \leq 2 \Rightarrow 2 \leq x \leq 6\)
\(
\begin{aligned}
& \therefore p(|x-4| \leq 2)=p(x=2)+p(x=3)+p(x=4) \\
& +p(x=5)+p(x=6) \\
& ={ }^8 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6+{ }^8 C_3\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^5+{ }^8 C_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^4 \\
& +{ }^8 C_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^3+{ }^8 C_6\left(\frac{1}{2}\right)^6\left(\frac{1}{2}\right)^2 \\
& =\frac{{ }^8 C_2+{ }^8 C_3+{ }^8 C_4+{ }^8 C_5+{ }^8 C_6}{2^8} \\
& =\frac{238}{256}=\frac{119}{128} \\
&
\end{aligned}
\)
A doctor is called to see a sick child. The doctor knows (prior to the visit) that \(90 \%\) of the sick children in that neighbourhood are sick with the flue, denoted by \(F\), while \(10 \%\) are sick with the measles, denoted by M. A well-known symptom of measles is a rash, denoted by \(R\). The probability of having a rash for a child sick with the measles is 0.95 . However, occasionally children with the flue also develop a rash with conditional probability 0.08 . Upon examination the child, the doctor finds a rash, then the probability that the child has the measles, is
(d)
\(
P(F)=0 \cdot 90, P(M)=0 \cdot 10 \text {, }
\)
\(
P\left(\frac{R}{F}\right)=0.08, P\left(\frac{R}{M}\right)=0.95
\)
\(
\begin{aligned}
\therefore \quad P\left(\frac{M}{R}\right) & =\frac{P(M) \cdot P\left(\frac{R}{M}\right)}{P(M) \cdot P\left(\frac{R}{M}\right)+P(F) \cdot P\left(\frac{R}{F}\right)} \\
& =\frac{0 \cdot 10 \times 0 \cdot 95}{0 \cdot 10 \times 0 \cdot 95+0 \cdot 90 \times 0 \cdot 08}=\frac{0 \cdot 095}{0 \cdot 167}=\frac{95}{167}
\end{aligned}
\)
Let \(\mathrm{N}\) denote the number that turns up when a fair die is rolled. If the probability that the system of equations
\(
\begin{aligned}
& x+y+z=1 \\
& 2 x+N y+2 z=2 \\
& 3 x+3 y+N z=3
\end{aligned}
\)
has unique solution is \(\frac{k}{6}\), then the sum of value of \(\mathrm{k}\) and all possible values of \(\mathrm{N}\) is
\(
\begin{aligned}
& x+y+z=1 \\
& 2 x+N y+2 z=2 \\
& 3 x+3 y+N z=3 \\
& \Delta=\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & N & 2 \\
3 & 3 & N
\end{array}\right| \\
& =(N-2)(N-3)
\end{aligned}
\)
For unique solution \(\Delta \neq 0\)
So \(\mathrm{N} \neq 2,3\)
\(\Rightarrow \mathrm{P}(\) system has unique solution \()=\frac{4}{6}\)
So \(\mathrm{k}=4\)
Therefore sum \(=4+1+4+5+6=20\)
Let \(\mathrm{M}\) be the maximum value of the product of two positive integers when their sum is 66 . Let the sample space \(S=\left\{x \in Z: x(66-x) \geq \frac{5}{9} M\right\}\) and the event \(A=\{x \in S: x\) is a multiple of 3\(\}\). Then \(\mathrm{P}(\mathrm{A})\) is equal to
\(
\begin{aligned}
& M=33 \times 33 \\
& x(66-x) \geq \frac{5}{9} \times 33 \times 33 \\
& 11 \leq x \leq 55 \\
& A:\{12,15,18, \ldots .54\} \\
& P(A)=\frac{15}{45}=\frac{1}{3}
\end{aligned}
\)
Let \(\mathrm{x}\) and \(\mathrm{y}\) be distinct integers where \(1 \leq \mathrm{x} \leq 25\) and \(1 \leq y \leq 25\). Then, the number of ways of choosing \(\mathrm{x}\) and \(\mathrm{y}\), such that \(\mathrm{x}+\mathrm{y}\) is divisible by 5 , is _____.
\(
x+y=5 \lambda
\)
Cases :
\(
\begin{array}{llc}
\mathbf{x} & \mathbf{y} & \text { Number of ways } \\
5 \lambda & 5 \lambda & 20 \\
5 \lambda+1 & 5 \lambda+4 & 25 \\
5 \lambda+2 & 5 \lambda+3 & 25 \\
5 \lambda+3 & 5 \lambda+2 & 25 \\
5 \lambda+4 & 5 \lambda+1 & 25
\end{array}
\)
Total = 120
Let \(\mathrm{N}\) be the sum of the numbers appeared when two fair dice are rolled and let the probability that \(\mathrm{N}-2, \sqrt{3 \mathrm{~N}}, \mathrm{~N}+2\) are in geometric progression be \(\frac{k}{48}\). Then the value of \(k\) is
\(
n(s)=36
\)
Given : \(\mathrm{N}-2, \sqrt{3 \mathrm{~N}}, \mathrm{~N}+2\) are in G.P.
\(
\begin{aligned}
& 3 N=(N-2)(N+2) \\
& 3 N=N^2-4 \\
& \Rightarrow N^2-3 N-4=0
\end{aligned}
\)
\(
(\mathrm{N}-4)(\mathrm{N}+1)=0 \Rightarrow \mathrm{N}=4 \text { or } \mathrm{N}=-1 \text { rejected }
\)
\(
\begin{aligned}
& (\text { Sum }=4) \equiv\{(1,3),(3,1),(2,2)\} \\
& n(A)=3
\end{aligned}
\)
\(
P(A)=\frac{3}{36}=\frac{1}{12}=\frac{4}{48} \Rightarrow k=4
\)
\(25 \%\) of the population are smokers. A smoker has 27 times more chances to develop lung cancer then a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is \(\frac{\mathrm{k}}{10}\). Then the value of \(\mathrm{k}\) is ____.
\(\mathrm{E}_1\) : Smokers
\(
\mathrm{P}\left(\mathrm{E}_1\right)=\frac{1}{4}
\)
\(\mathrm{E}_2\) : non-smokers
\(
\mathrm{P}\left(\mathrm{E}_2\right)=\frac{3}{4}
\)
\(
\mathrm{E} \text { : diagnosed with lung cancer }
\)
\(
\mathrm{P}\left(\mathrm{E} / \mathrm{E}_1\right)=\frac{27}{28}
\)
\(
\mathrm{P}\left(\mathrm{E} / \mathrm{E}_2\right)=\frac{1}{28}
\)
\(
P\left(E_1 / E\right)=\frac{P\left(E_1\right) P\left(E / E_1\right)}{P(E)}
\)
\(
=\frac{\frac{1}{4} \times \frac{27}{28}}{\frac{1}{4} \times \frac{27}{28}+\frac{3}{4} \times \frac{1}{28}}=\frac{9}{10}
\)
\(
K=9
\)
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the \(\mathrm{T}\)-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is
Required probability
\(
=1-\frac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}
\)
Taking \(\mathrm{D}_{(15)}\) as \(\frac{15 !}{e}\)
\(
\mathrm{D}_{(14)} \text { as } \frac{14 \text { ! }}{e}
\)
\(\mathrm{D}_{(13)}\) as \(\frac{13 !}{e}\)
We get, \(1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right)\)
\(
=1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08
\)
Alternate:
Given: Fifteen football players of a club-team are given \(15 \mathrm{t}\)-shirts with their names written on the backside.
To find: If the players pick up the t-shirts randomly, then the probability that at least 3 players pick the correct t-shirt is?
Solution:
We are said that fifteen football players of a club-team are given \(15 \mathrm{t}\)-shirts with their names written on the backside and the players pick up the t-shirts randomly. Then the probability that at least 3 players pick the correct t-shirt will be solved as follows:
Required Probability \(=1-\left(\mathrm{D}_{15}+{ }^{15} \mathrm{C}_1 \mathrm{D}_{14}+{ }^{15} \mathrm{C}_2 \mathrm{D}_{13}\right) / 15\) !
Now taking,
\(D_{15}\) as (15!/e), \(\quad D_{14}\) as (14!/e) and \(D_{13}\) as (13!/e)
We get,
\(
\begin{aligned}
\text { Required Probability } & =1-[15 ! / e+(15.14 !) / e+(15.14 .13 !) / 2 e] / 15 ! \\
& =1-(1 / e+1 / e+1 / 2 e) \\
& =1-(2+2+1) / 2 e \\
& =1-5 / 2 e \\
& =0.08
\end{aligned}
\)
\(\therefore\) Required Probability \(=0.08\)
Let \(S=\left\{w_1, w_2, \ldots.\right\}\) be the sample space associated to a random experiment. Let \(\mathrm{P}\left(\mathrm{w}_{\mathrm{n}}\right)=\frac{\mathrm{P}\left(\mathrm{w}_{\mathrm{n}-1}\right)}{2}, n \geq 2\).
Let \(\mathrm{A}=\{2 \mathrm{k}+3 \ell ; \mathrm{k}, \ell \in \mathbb{N}\}\) and \(\mathrm{B}=\left\{\mathrm{w}_{\mathrm{n}} ; \mathrm{n} \in \mathrm{A}\right\}\).
Then \(\mathrm{P}(\mathrm{B})\) is equal to
Let \(\mathrm{P}\left(\mathrm{w}_1\right)=\lambda\) then \(\mathrm{P}\left(\mathrm{w}_2\right)=\frac{\lambda}{2} \ldots \mathrm{P}\left(\mathrm{w}_{\mathrm{n}}\right)=\frac{\lambda}{2^{\mathrm{n}-1}}\)
As \(\sum_{\mathrm{k}=1}^{\infty} \mathrm{P}\left(\mathrm{w}_{\mathrm{k}}\right)=1 \Rightarrow \frac{\lambda}{1-\frac{1}{2}}=1 \Rightarrow \lambda=\frac{1}{2}\)
\(
\text { So, } P\left(w_n\right)=\frac{1}{2^n}
\)
\(
\mathrm{A}=\{2 \mathrm{k}+3 \ell ; \mathrm{k}, \ell \in \mathbb{N}\}=\{5,7,8,9,10 \ldots \ldots\}
\)
\(
\begin{aligned}
& \mathrm{B}=\left\{\mathrm{w}_{\mathrm{n}}: \mathrm{n} \in \mathrm{A}\right\} \\
& \mathrm{B}=\left\{\mathrm{w}_5, \mathrm{w}_7, \mathrm{w}_8, \mathrm{w}_9, \mathrm{w}_{10}, \mathrm{w}_{11}, \ldots\right\}
\end{aligned}
\)
\(
\begin{aligned}
& A=\mathbb{N}-\{1,2,3,4,6\} \\
& \therefore P(B)=1-\left[P\left(w_1\right)+P\left(w_2\right)+P\left(w_3\right)+P\left(w_4\right)+P\left(w_6\right)\right]
\end{aligned}
\)
\(
\begin{aligned}
& =1-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{64}\right] \\
& =1-\frac{32+16+8+4+1}{64}=\frac{3}{64}
\end{aligned}
\)
If an unbiased die, marked with \(-2,-1,0,1,2,3\) on its faces, is through five times, then the probability that the product of the outcomes is positive, is :
Either all outcomes are positive or any two are negative.
Now, \(\mathrm{p}=\mathrm{P}(\) positive \()=\frac{3}{6}=\frac{1}{2}\)
\(
\mathrm{q}=\mathrm{p}(\text { negative })=\frac{2}{6}=\frac{1}{3}
\)
Required probability
\(
\begin{gathered}
={ }^5 \mathrm{C}_5\left(\frac{1}{2}\right)^5+{ }^5 \mathrm{C}_2\left(\frac{1}{3}\right)^2\left(\frac{1}{2}\right)^3+{ }^5 \mathrm{C}_4\left(\frac{1}{3}\right)^4\left(\frac{1}{2}\right)^1 \\
=\frac{521}{2592}
\end{gathered}
\)
A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is p. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is \(q\). If \(\mathrm{p}: \mathrm{q}=\mathrm{m}\) \(: n\), where \(m\) and \(n\) are coprime, then \(m+n\) is equal to
\(
\begin{aligned}
& \mathrm{p}=\frac{{ }^6 \mathrm{C}_1}{6 \times 6}=\frac{1}{6} \\
& \mathrm{q}=\frac{{ }^6 \mathrm{C}_1 \times{ }^5 \mathrm{C}_1 \times 4}{6 \times 6 \times 6 \times 6}=\frac{5}{54} \\
& \therefore \mathrm{p}: \mathrm{q}=9: 5 \Rightarrow \mathrm{m}+\mathrm{n}=14
\end{aligned}
\)
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is
\(
\begin{aligned}
& \frac{{ }^5 \mathrm{C}_2+{ }^6 \mathrm{C}_2}{{ }^2 \mathrm{C}_2+{ }^3 \mathrm{C}_2+{ }^4 \mathrm{C}_2+{ }^5 \mathrm{C}_2+{ }^6 \mathrm{C}_2}=\frac{10+15}{1+3+6+10+15} \\
& =\frac{25}{35}=\frac{5}{7}
\end{aligned}
\)
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