NCERT Exemplar MCQs

Hint

\(
\text { The outcomes in the sample space } S \text { are } 52 \text { cards in the deck. }
\)

Hint

Let E be the event that a black face card is chosen. The outcomes in E are Jack, Queen, King or spades or clubs. Symbolically \(E=\{J, Q, K \text {, of spades and clubs }\}\)

Hint

All possible genders are expressed as :
\( S =\{ BBB , BBG , BGB , BGG , GBB , GBG , GGB , GGG \} \)

Hint

(i)Let A denote the event: ‘exactly one child is a girl’
\(
\begin{aligned}
& A =\{ BBG , BGB , GBB \} \\
& P ( A )=\frac{3}{8}
\end{aligned}
\)

(ii) Let B denote the event that at least two children are girls.
\(
B=\{ GGB , GBG , BGG , GGG \}, P ( B )=\frac{4}{8} .
\)

(iii) Let C denote the event : ‘no child is a girl’.
\(
\begin{aligned}
& C=\{B B B\} \\
\therefore \quad & P(C)=\frac{1}{8}
\end{aligned}
\)

Hint

2 digit positive integers which are multiples of 3 are \(12,15,18, \ldots, 99\). Thus, there are 30 such integers.

Hint

2-digit positive integers are \(10,11,12, \ldots, 99\). Thus, there are 90 such numbers. Since out of these, 30 numbers are multiple of 3 , therefore, the probability that a randomly chosen positive 2-digit integer is a multiple of 3 , is \(\frac{30}{90}=\frac{1}{3}\).

Hint

A PIN is a sequence of four symbols selected from 36 ( 26 letters +10 digits) symbols.
By the fundamental principle of counting, there are \(36 \times 36 \times 36 \times 36=36^4=1,679,616\) PINs in all. When repetition is not allowed the multiplication rule can be applied to conclude that there are
\(
36 \times 35 \times 34 \times 33=1,413,720 \text { different PINs }
\)
The number of PINs that contain at least one repeated symbol
\(
=1,679,616-1,413,720=2,65,896
\)
Thus, the probability that a randomly chosen PIN contains a repeated symbol is
\(
\frac{265,896}{1,679,616}=.1583
\)

Hint

\(
\text { Since } P ( D )=-0.20 \text {, this is not possible as } 0 \leq P ( A ) \leq 1 \text { for any event } A \text {. }
\)

Hint

\(P ( S )= P ( A \cup B \cup C \cup D )=\frac{9}{120}+\frac{45}{120}+\frac{27}{120}+\frac{46}{120}=\frac{127}{120} \neq 1\).

This violates the condition that \(P ( S )=1\).

Hint

Let B be the event that a truck stopped at the roadblock will have faulty brakes and T be the event that it will have badly worn tires. We have \(P ( B )=0.23\),
\(
\begin{aligned}
& P ( T )=0.24 \text { and } P ( B \cup T )=0.38 \\
& \text { and } P ( B \cup T )= P ( B )+ P ( T )- P ( B \cap T ) \\
& \text { So } 0.38=0.23+0.24- P ( B \cap T ) \\
& \Rightarrow P ( B \cap T )=0.23+0.24-0.38=0.09
\end{aligned}
\)

Hint

Let C be the event that the person will have his teeth cleaned and F and E be the event of getting cavity filled or tooth extracted, respectively. We are given
\(
\begin{aligned}
P ( C ) & =0.48, P ( F )=0.25, \quad P ( E )=.20, \quad P ( C \cap F )=.09, \\
P ( C \cap E ) & =0.12, P ( E \cap F )=0.07 \text { and } P ( C \cap F \cap E )=0.03
\end{aligned}
\)
Now,
\(
\begin{aligned}
P ( C \cup F \cup E )= & P ( C )+ P ( F )+ P ( E ) \\
& – P ( C \cap F )- P ( C \cap E )- P ( F \cap E ) \\
& + P ( C \cap F \cap E ) \\
= & 0.48+0.25+0.20-0.09-0.12-0.07+0.03 \\
= & 0.68
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
P \text { (Blue or White) } & = P \text { (Blue) }+ P \text { (White) } \\
& =\frac{10}{80}+\frac{20}{80}=\frac{30}{80}=\frac{3}{8}
\end{aligned}
\)
(b)
\(
\begin{aligned}
& P \text { (numbered } 1,2,3,4 \text { or } 5) \\
& = P (1 \text { of any colour })+ P (2 \text { of any colour }) \\
& \quad+ P (3 \text { of any colour })+ P (4 \text { of any colour })+ P (5 \text { of any colour })
\end{aligned}
\)
\(
=\frac{4}{80}+\frac{4}{80}+\frac{4}{80}+\frac{4}{80}+\frac{4}{80}=\frac{20}{80}=\frac{2}{8}=\frac{1}{4}
\)
(c) \(P\) (Red or yellow and numbered \(1,2,3\) or 4 )
\(= P (\) Red numbered \(1,2,3\) or 4\()+ P (\) yellow numbered \(1,2,3\) or 4\()\)
\(
=\frac{4}{80}+\frac{4}{80}=\frac{8}{80}=\frac{1}{10}
\)
(d)
\(
\begin{aligned}
& P (\text { numbered } 5,15,25 \text { or } 35) \\
& = P (5)+ P (15)+ P (25)+ P (35) \\
& = P (5 \text { of White, Red, Yellow, Blue })+ P (15 \text { of White, Yellow })+ P (25 \text { of Yellow }) \\
& \quad+ P (35 \text { of Yellow })
\end{aligned}
\)
\(
=\frac{4}{80}+\frac{2}{80}+\frac{1}{80}+\frac{1}{80}=\frac{8}{80}=\frac{1}{10}
\)
(e)
\(P\) (White and numbered higher than 12 or Yellow and numbered higher than 26 )
\(= P\) (White and numbered higher than 12 ) + \(P\) (Yellow and numbered higher than 26 )

Hint

Since a leap year has 366 days and hence 52 weeks and 2 days. The 2 days can be SM, MT, TW, WTh, ThF, FSt, StS.
Therefore, \(P(53\) Sundays or 53 Mondays \()=\frac{3}{7}\).

Hint

Since a 3-digit number cannot start with digit 0 , the hundredth place can have any of the 4 digits. Now, the tens and units place can have all the 5 digits. Therefore, the total possible 3-digit numbers are \(4 \times 5 \times 5\), i.e., 100. The total possible 3 digit numbers having all digits same \(=4\) Hence, P (3-digit number with same digits) \(=\frac{4}{100}=\frac{1}{25}\).

Hint

In a chess board, there are 64 squares of which 32 are white and 32 are black. Since 2 of one colour and 1 of other can be \(2 W, 1 B\), or \(1 W, 2 B\), the number of ways is \(\left({ }^{32} C _2 \times{ }^{32} C _1\right) \times 2\) and also, the number of ways of choosing any 3 boxes is \({ }^{64} C _3\).
Hence, the required probability \(=\frac{{ }^{32} C _2 \times{ }^{32} C _1 \times 2}{{ }^{64} C _3}=\frac{16}{21}\).

Hint

\(
\text { We have } P ( A \cup B )=\frac{1}{2}
\)
\(
\begin{aligned}
& \Rightarrow P ( A \cup( B – A ))=\frac{1}{2} \\
& \Rightarrow P ( A )+ P ( B – A )=\frac{1}{2} \text { (since } A \text { and } B – A \text { are mutually exclusive) } \\
& \Rightarrow 1- P (\overline{ A })+ P ( B – A )=\frac{1}{2} \\
& \Rightarrow 1-\frac{2}{3}+ P ( B – A )=\frac{1}{2}
\end{aligned}
\)
\(
\begin{aligned}
& \Rightarrow P(B-A)=\frac{1}{6} \\
& \Rightarrow P(\bar{A} \cap B)=\frac{1}{6} \quad \text { (since } \bar{A} \cap B \equiv B-A \text { ) }
\end{aligned}
\)

Hint

ABCDEF is a regular hexagon. Total number of triangles \({ }^6 C _3=20\). (Since no three points are collinear). Of these only \(\triangle ACE ; \triangle BDF\) are equilateral triangles.
Therefore, required probability \(=\frac{2}{20}=\frac{1}{10}\).

Hint

Let \(3 P ( A )=2 P ( B )= P ( C )=p\) which gives \(p(A)\) \(=\frac{p}{3}, P ( B )=\frac{p}{2}\) and \(P ( C )=p\)
Now since \(A , B , C\) are mutually exclusive and exhaustive events, we have
\(
P ( A )+ P ( B )+ P ( C )=1
\)
\(
\Rightarrow \quad \frac{p}{3}+\frac{p}{2}+p=1 \quad \Rightarrow \quad p=\frac{6}{11}
\)
Hence, \(P ( A )=\frac{p}{3}=\frac{2}{11}\)

Hint

Total number of mappings from a set A having \(n\) elements onto itself is \(n^n\)
Now, for one to one mapping the first element in A can have any of the \(n\) images in A; the \(2^{\text {nd }}\) element in A can have any of the remaining \((n-1)\) images, counting like this, the \(n^{\text {th }}\) element in A can have only 1 image.

Therefore, the total number of one to one mappings is \({n!}\).
\(
\text { Hence the required probability is } \frac{n!}{n^n}=\frac{n (n-1)!}{n n^{n-1}}=\frac{(n-1)!}{n^{n-1}} \text {. }
\)

Hint

Word ALGORITHM has 9 letters.
If GOR remain together, then it will remain together.
\(\therefore\) Number of letters \(=\) AL GOR ITHM \(=6+1=7\)
Number of words \(=7!\)
and the total number of words from ALGORITHM \(=9\) !
So, the required probability \(=\frac{7!}{9!}=\frac{7!}{9 \cdot 8 \cdot 7!}=\frac{1}{72}\)
Hence, the required probability \(=\frac{1}{72}\).

Hint

Number of desk occupied by one couple \(=1\)
Only \((4+1)=5\) persons to be assigned.
\(\therefore\) Number of ways of assigning these 5 persons
\(
=5!\times 2 \text { ! }
\)
Total number of ways of assigning 6 persons \(=6!\)
\(\therefore\) Probability that a couple has adjacent desk
\(
=\frac{5!\times 2!}{6!}=\frac{1}{3}
\)
So, the probability that the married couple will have no-adjacent desks \(=1-\frac{1}{3}=\frac{2}{3}\).
Hence, the required probability \(=\frac{2}{3}\).

Hint

We have multiples of 2 , from 1 to 1000 are
\(
2,4,6,8, \ldots, 1000
\)
Let \(n\) be the number of terms
\(
a=2, d=2, a_n=1000
\)
\(
\begin{aligned}
a_n & =a+(n-1) d \\
1000 & =2+(n-1) 2 \\
1000 & =2 n \quad \Rightarrow \quad n=500
\end{aligned}
\)
Now multiple of 9 from 1 to 1000 are \(9,18,27, \ldots, 999\)
Here \(a=9, d=9\) and \(a_m=999 \quad\) [ \(m\) is the number of terms]
\(
\begin{aligned}
& & a_m & =a+(m-1) d \\
\Rightarrow & & 999 & =9+(m-1) 9 \\
\Rightarrow & & 999 & =9 m \Rightarrow m=111
\end{aligned}
\)
Now number multiples of 2 and 9 are
\(
18,36,54, \ldots, 990
\)
Here \(a=18, a_p=990, d=18\) [ \(p\) is the number of terms]
\(
\begin{aligned}
& \therefore \quad a_p=a+(p-1) d \\
& 990=18+(p-1) 18 \\
& \Rightarrow \quad 990=18+18 p-18 \\
& \Rightarrow \quad 18 p=990 \quad \Rightarrow \quad p=55 \\
& \therefore \text { Number of multiples of } 2 \text { or } 9=500+111-55 \\
& =556 \\
& \therefore \text { Required probability }=\frac{ P ( E )}{n( E )}=\frac{556}{1000}=0.556 \\
&
\end{aligned}
\)
Hence, the required probability \(=0.556\).

Hint

Number of sample space \(=6\)
(i) Given that 2 appears on the \(k\) th roll of the die.
So first \((k-1)\) th roll have 5 out comes each and \(k\) th roll results 2 i.e. only 1 outcome.
\(\therefore\) Number of element of sample space correspond to the event that 2 appears on the \(k\) th roll of the die \(=5^{k-1}\).
(ii) In this case, 2 appears not later than \(k\) th roll of the die, then it is possible that 2 comes in first throw i.e. 1 outcome. If 2 does not appear in first throw, then outcomes will be 5 and 2 outcomes in second throw i.e. 1 outcome.
\(\therefore \quad\) Possible out come \(=5 \times 1=5\)
Similarly, if \(2\) does not appear in second throw and appears in third throw
\(\therefore \quad\) Possible outcome \(=5 \times 5 \times 1\)
Now we have the series:
\(
\begin{aligned}
& =1+5+5 \times 5+5 \times 5 \times 5+\ldots+5^{k-1} \\
& =1+5+5^2+5^3+\ldots+5^{k-1}
\end{aligned}
\)
\(
\begin{aligned}
=\frac{1 .\left(r^k-1\right)}{r-1} & =\frac{5^k-1}{5-1} \\
& =\frac{5^k-1}{4}
\end{aligned}
\)
Hence, the required answer \(=\frac{5^k-1}{4}\).

Hint

Given that probability of even numbers
\(
=\frac{1}{2} \times \text { Probability of odd numbers }
\)
\(
\begin{array}{lr}
\Rightarrow & P \text { (Odd }): P (\text { Even })=2: 1 \\
\therefore & P \text { (odd number })=\frac{2}{2+1}=\frac{2}{3} \\
\text { and } & P \text { (even number })=\frac{1}{2+1}=\frac{1}{3}
\end{array}
\)
Also given that, G the event that a number greater than 3 occurs in a single throw of die.
\(\therefore\) The possible outcome are 4,5 and 6 out of which two are even and one is odd.
\(
\begin{aligned}
\therefore \text { Required probability } & = P ( G ) \\
& =2 \times P (\text { even }) \times P \text { (odd }) \\
& =2 \times \frac{1}{3} \times \frac{2}{3}=\frac{4}{9}
\end{aligned}
\)
Hence, the required probability is \(\frac{4}{9}\).

Hint

Let \(E_1\) be the event that a family owns colour television set \(E _2\) be the event that the family owns black and white television set.
Given that
\(
\begin{aligned}
P \left( E _1\right) & =0.87, P \left( E _2\right)=0.36 \\
P \left( E _1 \cap E _2\right) & =0.30
\end{aligned}
\)
\(\therefore\) The probability that a family owns either colour television set or black and white television set
\(
\begin{aligned}
\therefore \quad P\left(E_1 \cup E_2\right) & =P\left(E_1\right)+P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\
& =0.87+0.36-0.30 \\
& =0.93
\end{aligned}
\)
Hence, the required probability \(=0.93\).

Hint

Given that \(P ( A )=0.35\) and \(P ( B )=0.45\)
Since the events A and B are mutually exclusive then \(P ( A \cap B )=0\)
(i) \(\quad P \left( A ^{\prime}\right)=1- P ( A )=1-0.35=0.65\)
(ii) \(\quad P \left( B ^{\prime}\right)=1- P ( B )=1-0.45=0.55\)
(iii) \(P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )\)
\(
\begin{aligned}
& =0.35+0.45-0 \\
& =0.80
\end{aligned}
\)
(iv) \(P ( A \cap B )=0\)
[ \(\because\) A and \(B\) are mutually exclusive events]
(v) \(P \left( A \cap B ^{\prime}\right)= P ( A )- P ( A \cap B )\)
\(
=0.35-0=0.35
\)
(vi) \(P \left( A ^{\prime} \cap B ^{\prime}\right)=1- P ( A \cup B )=1-0.80=0.20\).

Hint

Let \(E_1, E_2, E_3, E_4\) and \(E_5\) be the events that the surgeries are rated as very complex, complex, routine, simple and very simple respectively.
\(
\begin{aligned}
& \therefore P \left( E _1\right)=0.15, P \left( E _2\right)=0.20, P \left( E _3\right)=0.31, P \left( E _4\right)=0.26 \\
& \text { and } P \left( E _5\right)=0.08
\end{aligned}
\)
\(
\begin{aligned}
& \text { (i) } \begin{aligned}
& P (\text { complex or very complex })= P \left( E _1 \text { or } E _2\right) \\
& \Rightarrow P \left( E _1 \cup E _2\right)= P \left( E _1\right)+ P \left( E _2\right)- P \left( E _1 \cap E _2\right) \\
&=0.15+0.20-0 \\
&=0.35 \quad[\because \text { All event are independent }]
\end{aligned}
\end{aligned}
\)
(ii) P (neither very complex nor Very simple \()= P \left( E _1^{\prime} \cap E _5^{\prime}\right)\)
\(
\begin{aligned}
& =P\left(E_1 \cup E_5\right)^{\prime}=1-P\left(E_1 \cup E_5\right) \\
& =1-\left[P\left(E_1\right)+P\left(E_5\right)\right] \\
& =1-[0.15+0.08]=1-0.23=0.77
\end{aligned}
\)
(iii) P (routine or complex \()= P \left( E _3\right.\) or \(\left.E _2\right)\)
\(
\begin{aligned}
& =P\left(E_3 \cup E_2\right)=P\left(E_3\right)+P\left(E_2\right) \\
& =0.31+0.20=0.51
\end{aligned}
\)
(iv) P (routine or simple \()= P \left( E _3 \cup E _4\right)= P \left( E _3\right)+ P \left( E _4\right)\)
\(
=0.31+0.26=0.57
\)

Hint

Given that A is twice as likely to be selected as B i.e. \(\quad P ( A )=2 P ( B )\) and C is twice as likely to be selected as D
\(
\begin{array}{llll}
\therefore & P ( C )=2 P ( D ) & \Rightarrow & P ( B )=2 P ( D ) \\
\Rightarrow & \frac{ P ( A )}{2}=2 P ( D ) & \Rightarrow & P ( D )=\frac{1}{4} P ( A )
\end{array}
\)
Now B and C are given about the same chance
\(
\begin{array}{lr}
\therefore & P ( B )= P ( C ) \\
\text { Since, } & \text { sum of all probabilities }=1 \\
\therefore & P ( A )+ P ( B )+ P ( C )+ P ( D )=1 \\
\Rightarrow & P ( A )+\frac{ P ( A )}{2}+\frac{ P ( A )}{2}+\frac{ P ( A )}{4}=1 \\
\Rightarrow & 4 P ( A )+2 P ( A )+2 P ( A )+ P ( A )=4
\end{array}
\)
\(
9 P ( A )=4 \Rightarrow P ( A )=\frac{4}{9}
\)
(i)
\(
\begin{aligned}
P(C \text { will be selected }) & =P(C)=P(B)=\frac{P(A)}{2} \\
& =\frac{4}{9} \times \frac{1}{2}=\frac{2}{9}
\end{aligned}
\)
(ii)
\(
\begin{aligned}
P ( A \text { will not be selected }) & = P \left( A ^{\prime}\right)=1- P ( A ) \\
& =1-\frac{4}{9}=\frac{5}{9}
\end{aligned}
\)

Hint

Let \(E _1, E _2, E _3\) and \(E _4\) be the events that John promoted, Rita promoted, Aslam promoted and Gurpreet promoted respectively.
\(
\therefore \quad \text { Sample space } S =\left\{ E _1, E _2, E _3, E _4\right\}
\)
Given that probability of John’s promotion is same as that of Gurpreet
\(
\therefore \quad P\left(E_1\right)=P\left(E_4\right)
\)
Rita’s chances of promotion are twice as likely as John
\(
\therefore \quad P \left( E _2\right)=2 P \left( E _1\right)
\)
and Aslam’s chances of promotion are 4 times that of John
\(
\therefore \quad P \left( E _3\right)=4 P \left( E _1\right)
\)
Since, the sum of all the probabilities \(=1\)
\(
\begin{aligned}
P \left( E _1\right)+ P \left( E _2\right)+ P \left( E _3\right)+ P \left( E _4\right) & =1 \\
P \left( E _1\right)+2 P \left( E _1\right)+4 P \left( E _1\right)+ P \left( E _1\right) & =1 \\
8 P \left( E _1\right) & =1 \\
P \left( E _1\right) & =\frac{1}{8}
\end{aligned}
\)
(i)
\(
\begin{aligned}
P (\text { John promoted }) & = P \left( E _1\right)=\frac{1}{8} \\
P (\text { Rita promoted }) & = P \left( E _2\right)=2 P \left( E _1\right)=2 \times \frac{1}{8}=\frac{1}{4} \\
P (\text { Aslam promoted }) & = P \left( E _3\right)=4 P \left( E _1\right)=4 \times \frac{1}{8}=\frac{1}{2} \\
P (\text { Gurpreet promoted }) & = P \left( E _4\right)= P \left( E _1\right)=\frac{1}{8}
\end{aligned}
\)
(ii)
\(
\begin{aligned}
& P (\text { John promoted or Gurpreet promoted })= P \left( E _1 \cup E _4\right) \\
& \Rightarrow \quad \begin{aligned}
P \left( E _1 \cup E _4\right) & = P \left( E _1\right)+ P \left( E _4\right)- P \left( E _1 \cap E _4\right) \\
& =\frac{1}{8}+\frac{1}{8}-0 \quad\left[\because P \left( E _1 \cap E _4\right)=0\right] \\
& =\frac{1}{4}
\end{aligned}
\end{aligned}
\)

Hint

From the given Venn diagram
(i) \(P ( A )=0.13+0.07=0.20\)
(ii) \(P ( B \cap \overline{ C })= P ( B )- P ( B \cap C )[latex] [latex]=0.07+0.10+0.15-0.15\)
\(=0.07+0.10\)
\(=0.17\)
(iii)
\(
\begin{aligned}
P ( A \cup B ) & = P ( A )+ P ( B )- P ( A \cap B ) \\
& =0.13+0.07+0.07+0.10+0.15-0.07 \\
& =0.13+0.07+0.10+0.15 \\
& =0.45
\end{aligned}
\)
(iv)
\(
\begin{aligned}
P ( A \cap \overline{ B }) & = P ( A )- P ( A \cap B ) \\
& =0.13+0.07-0.07=0.13
\end{aligned}
\)
(v) \(P ( B \cap C )=0.15\)
(vi) P (exactly one of the three occurs) \(=0.13+0.10+0.28\) \(=0.51\)

Hint

Given that one of the two urns is choosen, then a ball is randomly choosen from the urn, then a second ball is choosen at random from the same urn without replacing the first ball
(i) Sample space \(S =\left\{ B _1 B_2, B_1 W, B _2 B_1, B_2 W, WB _1, WB _2, BW _1\right.\), \(\left.BW _2, W_1 B, W _1 W_2, W_2 B, W _2 W_1\right\}\)
Total number of Sample space, S \(=12\)
(ii) If two black balls are choosen then the favourable events are \(B_1 B_2, B_2 B_1\) i.e. 2
\(\therefore\) Required probability \(=\frac{2}{12}=\frac{1}{6}\)
(iii) If two balls of opposite colours are choosen then, the required probability \(=\frac{8}{12}=\frac{2}{3}\)

Hint

Given that: \(\quad\) Number of red balls \(=8\)
Number of white balls \(=5\)
(i) P (all the three balls are white)
\(
\begin{aligned}
& =\frac{{ }^5 C_3}{{ }^{13} C_3}=\frac{\frac{5!}{3!2!}}{\frac{13!}{3!10!}}=\frac{5!}{3!2!} \times \frac{3!10!}{13!} \\
& =\frac{5!}{2!} \times \frac{10!}{13 \times 12 \times 11 \times 10!} \\
& =\frac{5 \times 4 \times 3 \times 2!}{2!} \times \frac{1}{13 \times 12 \times 11} \\
& =\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{143}
\end{aligned}
\)
(ii) P (all the three balls are red)
\(
=\frac{{ }^8 C_3}{{ }^{13} C_3}=\frac{\frac{8!}{3!5!}}{\frac{13!}{3!10!}}=\frac{8!}{3!5!} \times \frac{3!10!}{13!}
\)
\(
\begin{aligned}
& =\frac{8 \times 7 \times 6 \times 5!}{5!} \times \frac{10!}{13 \times 12 \times 11 \times 10!} \\
& =\frac{8 \times 7 \times 6}{13 \times 12 \times 11}=\frac{28}{143}
\end{aligned}
\)
(iii) P (one ball is red and two balls are white)
\(
\begin{aligned}
& =\frac{{ }^8 C_1 \times{ }^5 C_2}{{ }^{13} C_3}=\frac{8 \times 10}{\frac{13!}{3!\times 10!}} \\
& =\frac{8 \times 10}{13!} \times \frac{3!\times 10!}{1}=\frac{8 \times 10 \times 3 \times 2 \times 10!}{13 \times 12 \times 11 \times 10!} \\
& =\frac{8 \times 10 \times 6}{13 \times 12 \times 11}=\frac{40}{143}
\end{aligned}
\)

Hint

Total number of word is ASSASSINATION are 13.
\(
\text { Where, we have } 3 A^{\prime} s , 4 S^{\prime}, 2 I ^{\prime} s , 2 N^{\prime} s , 1 T^{\prime} s \text { and } 1 O ^{\prime} s \text {. }
\)
(i) If 4 S’s come consecutively in the word, then arrangement may be as follows:
S S S S A A A IINNTO
1 Group 9 others
\(\therefore\) Number of words when all S’s are together \(=\frac{10!}{3!2!2!}\) and the total number of word formed from the words
\(
\text { ASSASSINATION }=\frac{13!}{3!4!2!2!}
\)
\(
\begin{aligned}
\therefore \text { Required probability } & =\frac{\frac{10!}{3!2!2!}}{\frac{13!}{3!4!2!2!}}=\frac{10!}{3!2!2!} \times \frac{3!4!2!2!}{13!} \\
& =\frac{10!4!}{13!}=\frac{10!\times 4 \times 3 \times 2}{13 \times 12 \times 11 \times 10!} \\
& =\frac{2}{143}
\end{aligned}
\)
(ii) If 2 I’s and 2 N ‘s come together then there are 10 alphabets Number of words when 2 I’s and 2 N ‘s are come together
\(
=\frac{10!}{3!4!} \times \frac{4!}{2!2!}
\)
\(
\begin{aligned}
\therefore \text { Required probability } & =\frac{\frac{10!}{3!4!} \times \frac{4!}{2!2!}}{\frac{13!}{3!4!2!2!}} \\
& =\frac{4!10!}{2!2!3!4!} \times \frac{3!4!2!2!}{13!} \\
& =\frac{4!10!}{13!}=\frac{4 \times 3 \times 2 \times 10!}{13 \times 12 \times 11 \times 10!} \\
& =\frac{4 \times 3 \times 2}{13 \times 12 \times 11}=\frac{2}{143}
\end{aligned}
\)
(iii) If all A’s are coming together, then three are 11 alphabets Number of words when all A’s come together
\(
=\frac{11!}{4!2!2!}
\)
\(\therefore\) Probability when all A’s come together
\(
\begin{aligned}
& =\frac{\frac{11!}{4!2!2!}}{\frac{13!}{4!3!2!2!}}=\frac{11!}{4!2!2!} \times \frac{4!3!2!2!}{13!} \\
& =\frac{11!\times 3!}{13!}=\frac{6}{13 \times 12}=\frac{1}{26}
\end{aligned}
\)
\(\therefore\) Required probability when all A’s do not come together
\(
=1-\frac{1}{26}=\frac{25}{26}
\)
(iv) If no two A’s are together, then arranging the alphabets except A’s
\(
– S – S – S – S – I – N – T – I – O – N –
\)
Number of ways of arranging all alphabets except A’s
\(
=\frac{10!}{4!2!2!}
\)
There are 11 vacant places between these alphabets.
\(\therefore 3\) A’s can be placed in 11 places in \({ }^{11} C _3\) ways
\(
=\frac{11!}{3!8!}
\)
\(\therefore\) Total number of words when no two A’s together
\(
\begin{aligned}
& =\frac{11!}{3!8!} \times \frac{10!}{4!2!2!} \\
\therefore \text { Required probability } & =\frac{11!\times 10!}{3!8!4!2!2!} \times \frac{4!3!2!2!}{13!} \\
& =\frac{10!}{8!\times 13 \times 12}=\frac{10 \times 9 \times 8!}{8!\times 13 \times 12} \\
& =\frac{10 \times 9}{13 \times 12}=\frac{15}{26}
\end{aligned}
\)

Hint

Total number of cards \(=52\)
Favourable events \(=4\) kings +13 hearts +26 red \(-13-2=28\)
\(\therefore\) Required probability \(=\frac{28}{52}=\frac{7}{13}\).

Hint

\(
\text { Given that: } \quad \begin{aligned}
S & =\left\{e_1, e_2, e_3, e_4, e_5, e_6, e_7, e_8, e_9\right\} \\
A & =\left\{e_1, e_5, e_8\right\} \text { and } B =\left\{e_2, e_5, e_8, e_9\right\}
\end{aligned}
\)
\(
\begin{aligned}
& P \left(e_1\right)= P \left(e_2\right)=0.08 \\
& P \left(e_3\right)= P \left(e_4\right)= P \left(e_5\right)=0.1 \\
& P \left(e_6\right)= P \left(e_7\right)=0.2, P \left(e_8\right)= P \left(e_9\right)=0.07
\end{aligned}
\)
(i)
\(
\begin{aligned}
P ( A ) & = P \left(e_1\right)+ P \left(e_5\right)+ P \left(e_8\right) \\
& =0.08+0.1+0.07=0.25
\end{aligned}
\)
(ii)
\(
\begin{aligned}
P ( A \cup B ) & = P ( A )+ P ( B )- P ( A \cap B ) \\
\text { But } P ( B ) & = P \left(e_2\right)+ P \left(e_5\right)+ P \left(e_8\right)+ P \left(e_9\right) \\
& =0.08+0.1+0.07+0.07 \\
& =0.32
\end{aligned}
\)
\(
\begin{aligned}
\text { and } P ( A \cap B ) & =\left\{e_5, e_8\right\} \\
& = P \left(e_5\right)+ P \left(e_8\right)=0.1+0.07=0.17
\end{aligned}
\)
Putting the values in eq. (i) we get
\(
\begin{aligned}
P ( A \cup B ) & =0.25+0.32-0.17 \\
& =0.40
\end{aligned}
\)
(iii)
\(
\begin{aligned}
A \cup B & =\left\{e_1, e_2, e_5, e_8, e_9\right\} \\
\therefore \quad P ( A \cup B ) & = P \left(e_1\right)+ P \left(e_2\right)+ P \left(e_5\right)+ P \left(e_8\right)+ P \left(e_9\right) \\
& =0.08+0.08+0.1+0.07+0.07=0.40
\end{aligned}
\)
(iv) \(\quad P (\overline{ B })=1- P ( B )=1-0.32=0.68\)

Hint

(i) Possible outcomes of a single throw of die \(S=\{1,2,3,4,5,6\}\) out of which \(1,3,5\) are odd
\(\therefore\) Required probability \(=\frac{3}{6}=\frac{1}{2}\)
(ii) When a fair coin is tossed twice, then the sample space
\(
S =\{ HH , HT , TH , TT \}
\)
\(\therefore\) Probability of getting atleast one head \(( HH , HT , TH )\)
\(
=\frac{3}{4}
\)
(iii) Favourable events are 4 kings +2 of hearts +3 of spades
\(
\begin{aligned}
& =4+1+1=6 \\
& =\frac{6}{52}=\frac{3}{26}
\end{aligned}
\)
(iv) When a pair of dice is rolled, then total number of sample space \(=36\) out of which \((1,5),(5,1),(2,4),(4,2)\) and \((3,3)\) are the favourable events
\(\therefore\) Required probability \(=\frac{5}{36}\).

Hint

There are 365 days in a non-leap year and there are 7 days in a week
\(
\therefore \quad 365 \div 7=52 \text { weeks }+1 \text { day }
\)
So, this day may be Tuesday or Wednesday.
So, the required probability \(=\frac{1}{7}\).

Hint

Set of three consecutive numbers from 1 to 20 are \(1,2,3\); \(2,3,4 ; 3,4,5 ; \ldots, 18,19,20\).
So, the probability that the numbers are consecutive
\(
\begin{aligned}
& =\frac{18}{{ }^{20} C_3}=\frac{18}{\frac{20!}{3!17!}}=\frac{18 \cdot 3!\cdot 17!}{20!} \\
& =\frac{18 \times 3 \times 2 \times 17!}{20 \times 19 \times 18 \times 17!}=\frac{3 \times 2}{20 \times 19}=\frac{3}{190}
\end{aligned}
\)
\(\therefore P\) (three numbers are not consecutive)
\(
=1-\frac{3}{190}=\frac{187}{190}
\)

Hint

We know that out of 52 playing cards 26 are of red and 26 are of black colour.
\(\therefore P\) (both cards of different colour)
\(
=\frac{26}{52} \times \frac{26}{51}+\frac{26}{52} \times \frac{26}{51}
\)
\(
=2 \times \frac{26}{52} \times \frac{26}{51}=\frac{26}{51}
\)

Hint

The two particular persons to be seated next each other then, they form one group.
Now the permutation of 6 persons \(=6!\times 2\) ! and Total number of permutations of 7 persons \(=7\) !
\(
\begin{aligned}
\therefore \text { Required probability } & =\frac{6!\times 2!}{7!} \\
& =\frac{6!\times 2}{7 \times 6!}=\frac{2}{7}
\end{aligned}
\)

Hint

Four digit number using the digits \(0,2,3,5\) with out repetition and divisible by 5 with the given conditions is
\(
\begin{array}{|l|l|l|l|}
\hline 3 & 2 & 1 & 1 \\
\hline
\end{array}
\)
If unit place be filled with 0
Then the number of ways \(=3 \times 2 \times 1 \times 1=6\)
\(
\begin{array}{|l|l|l|l|}
\hline 2 & 2 & 1 & 1 \\
\hline
\end{array}
\)
If unit place be filled with 5
Then the number of ways \(=2 \times 2 \times 1 \times 1=4\)
\(\therefore\) Total number of ways \(=6+4=10\)
Total number of ways of arranging the digits \(0,2,3,5\) to form 4-digit numbers without repetition is \(3 \times 3 \times 2 \times 1=18\)
\(\therefore\) Required probability \(=\frac{10}{18}=\frac{5}{9}\)

Hint

For mutually exclusive events,
\(
P ( A \cap B )=0
\)
\(
\begin{aligned}
P ( A \cup B ) & = P ( A )+ P ( B ) & & {[\because P ( A \cap B )=0] } \\
P ( A )+ P ( B ) & \leq 1 & & \\
P ( A )+1- P (\overline{ B }) & \leq 1 & & {[ P ( B )=1- P (\overline{ B })] } \\
P ( A )- P (\overline{ B }) & \leq 0 & & \\
P ( A ) & \leq P (\overline{ B }) & &
\end{aligned}
\)

Hint

Given that: \(\quad P ( A \cup B )= P ( A \cap B )\)
\(
P ( A )+ P ( B )- P ( A \cap B )= P ( A \cap B )
\)
\(
\begin{aligned}
{[ P ( A )- P ( A \cap B )]+[ P ( B )- P ( A \cap B )] } & =0 \\
P ( A )- P ( A \cap B ) & \geq 0 \dots(i)
\end{aligned}
\)
\(
[\because P ( A \cap B ) \leq P ( A ) \text { or } P ( B )]
\)
and \(P(B)-P(A \cap B) \geq 0 \dots(ii)\)
From eq. (i) and (ii) we get
\(
P ( A )= P ( B )
\)

Hint

If all the girls sit together, then we consider it as 1 group
\(
\begin{array}{|l|l|l|l||l|l||l|}
\hline G & G & G & G & G & G & \\
\hline
\end{array}
\)
\(
\begin{aligned}
& \therefore \text { Total number of arrangement of } 6+1=7 \text { persons in a row } \\
& =7!\text { and the girls also interchanged their places with } 6!\text { ways. } \\
& \begin{aligned}
\therefore \text { Required probability } & =\frac{6!7!}{12!}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 7!}{12 \times 11 \times 10 \times 9 \times 8 \times 7!} \\
& =\frac{1}{132}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { Total number of alphabets in probability }=11 \\
& \quad \text { Number of vowels }=4 \\
& \therefore \quad \text { Required probability }=\frac{4}{11}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { given that: } & P(A \text { fails })=0.2 \\
& P(B \text { fails })=0.3
\end{array}
\)
\(
\begin{aligned}
\therefore \quad P \text { (either A or B fails }) & \leq P ( A \text { fails })+ P ( B \text { fails }) \\
& \leq 0.2+0.3 \\
& \leq 0.5
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
\text { Given that: } & P ( A \cup B )=0.6 \\
& P ( A \cap B )=0.2 \\
\therefore & P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )
\end{array}
\)
\(
\begin{aligned}
& \Rightarrow \quad 0.6= P ( A )+ P ( B )-0.2 \\
& \Rightarrow \quad P ( A )+ P ( B )=0.6+0.2=0.8 \\
& \text { and } \quad 1- P (\overline{ A })+1- P (\overline{ B })=0.8 \\
& \Rightarrow \quad P (\overline{ A })+ P (\overline{ B })=2-0.8=1.2 \\
&
\end{aligned}
\)

Hint

If M and N are any two events, then
\(
P ( M \cup N )= P ( M )+ P ( N )- P ( M \cap N )
\)

Hint

Given that:
\(
\begin{aligned}
P (\text { to see giraffee }) & =0.72 \\
P (\text { to see bears }) & =0.84 \\
P (\text { to see both giraffee and bears }) & =0.52
\end{aligned}
\)
\(
\begin{aligned}
& P \text { (to see giraffee or bear }) \\
&= P \text { (to see giraffee })+ P \text { (to see bear }) -P \text { (to see both) } \\
&=0.72+0.84-0.52 \\
&=1.04 \text { which is not possible. }
\end{aligned}
\)
Hence, the given statement is False.

Hint

Let \(E\) be the event that the student will pass and F be the event that he will get compartment
\(
\begin{aligned}
& \therefore P ( E )=0.73, P ( F )=0.13 \text { and } P ( E \cup F )=0.96 \\
& \therefore \quad P ( E \cup F )= P ( E )+ P ( F )- P ( E \cap F ) \\
& =0.73+0.13-0 \quad[\because P ( E \cap F )=0] \\
& =0.86 \\
& \text { But } \quad P ( E \cup F )=0.96 \\
&
\end{aligned}
\)
Hence, the given statement is False.

Hint

\(
\begin{aligned}
& \text { Sum of all probabilities }=1 \\
& \therefore P (0)+ P (1)+ P (2)+ P (3)+ P (4)+ P (5) \\
& =0.12+0.25+0.36+0.14+0.08+0.11 \\
& =1.06>1 \\
&
\end{aligned}
\)
Hence, the given statement is False.

Hint

It is given that
\(
\begin{aligned}
& P(A)=0.5 \\
& P(A \cap B) \leq 0.3 \\
& {[P(A \text { and } B)=P(A \cap B)]} \\
& \text { Now } P(A)+P(B)-P(A \cap B)=P(A \cup B) \leq 1 \\
& 0.5+P(B)-P(A \cap B) \leq 1 \\
& P(B) \leq 0.8
\end{aligned}
\)
we can say that, the probability of the event that \(B\) is selected can be at most 0.8
Hence, the given statement is True.

Hint

Here \(\quad P ( A \cap B ) \leq P ( A )\)
Which is always true.
Hence, given statement is True.

Hint

\(
\begin{aligned}
& \text { Here, } P ( A )=0.7, P ( B )=0.3 \\
& \begin{aligned}
P ( A \cap B ) & = P ( A ) \times P ( B ) \\
& =0.7 \times 0.3=0.21
\end{aligned}
\end{aligned}
\)
But the given probability is 0.4 .
Hence, the given statement is False.

Hint

Since, the two given events are not related to the same Sample space.
\(\therefore\) The sum of probabilities of two students getting distinction in their final examinations may be 1.2.
Hence, the given statement is True.

Hint

\(
\begin{aligned}
P \text { (loosing the game }) & =1-(0.77+0.08) \\
& =1-0.85=0.15
\end{aligned}
\)
Hence, the value of the filler is \(0 . 1 5\).

Hint

We know that the sum of all probabilities \(=1\)
\(
\therefore \quad P \left(e_1\right)+ P \left(e_2\right)+ P \left(e_3\right)+ P \left(e_4\right)=1
\)
\(
\begin{aligned}
0.1+0.5+0.1+ P \left(e_4\right) & =1 \\
0.7+ P \left(e_4\right) & =1 \\
P \left(e_4\right) & =1-0.7=0.3
\end{aligned}
\)
Hence, the value of the filler is 0.3.

Hint

Given that
\(
\begin{aligned}
& S=\{1,2,3,4,5,6\} \\
& E=\{1,3,5\} \\
& \therefore \quad \overline{ E }= S – E =\{2,4,6\} \\
&
\end{aligned}
\)
Hence, the value of the filler is \(\{2,4,6\}\).

Hint

Given that: \(P ( A )=0.3, P ( B )=0.2\)
\(
\begin{array}{rlrl}
& & P ( A \cap B ) & =0.1 \\
& & P ( A \cap \overline{ B }) & = P ( A )- P ( A \cap B ) \\
& & =0.3-0.1=0.2
\end{array}
\)
Hence, the value of the filler is \(0 . 2\).

Hint

(i) \(0.95=\) Very likely to happen, so it is close to 1 .
(ii) \(0.02=\) Very little chance of happening as the probability is very low.
(iii) \(-0.3=\) an incorrect assignment because probability is never negative.
(iv) \(0.5=\) as much chance of happening as not because sum of chances of happening and not happening is one.
(v) \(0=\) no chance of happening.

Hence, \((a) \leftrightarrow(i v),(b) \leftrightarrow(v),(c) \leftrightarrow(i),(d) \leftrightarrow(i i i),(e) \leftrightarrow(i i)\)

Hint

(a) If \(E _1\) and \(E _2\) are mutually exclusive events, then \(E _1 \cap E _2\) \(=\phi\).
(b) If \(E _1\) and \(E _2\) are mutually exclusive and exhaustive events then \(E_1 \cap E_2=\phi\) and \(E_1 \cup E_2=S\).
(c) If \(E _1\) and \(E _2\) have common outcomes, then \(\left(E_1-E_2\right) \cup\left(E_1 \cap E_2\right)=E_1\)
(d) If \(E _1\) and \(E _2\) are two events such that \(E _1 \subset E _2 \Rightarrow E _1 \cap E _2= E _1\)
\(
\text { Hence, }(a) \leftrightarrow(i v),(b) \leftrightarrow(i i i),(c) \leftrightarrow(i i),(d) \leftrightarrow(i)
\)