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In a simultaneous throw of two coins, the probability of getting at least one head is
Here \(S=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}\).
Let \(E=\) event of getting at least one head \(=\{H T, T H\), \(\mathrm{HH}\}\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{3}{4}
\)
Three unbiased coins are tossed. What is the probability of getting at least 2 heads?
Here \(S=\{\mathrm{TTT}, \mathrm{TTH}\), THT, \(\mathrm{HTT}\), THH, \(\mathrm{HTH}, \mathrm{HHT}, \mathrm{HHH}\}\). Let \(E=\) event of getting at least two heads \(=\{\mathrm{THH}, \mathrm{HTH}\), \(\mathrm{HHT}, \mathrm{HHH}\}\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{4}{8}=\frac{1}{2}
\)
Three unbiased coins are tossed. What is the probability of getting at most two heads?
Here \(S=\{\mathrm{TTT}, \mathrm{TTH}\), THT, HTT \(, \mathrm{THH}, \mathrm{HTH}, \mathrm{HHT}, \mathrm{HHH}\}\).
Let \(E=\) event of getting at most two heads.
Then, \(E=\{\mathrm{TTT}\), TTH, THT, HTT, THH, HTH, HHT \(\}\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{7}{8}
\)
In a single throw of a die, what is the probability of getting a number greater than 4?
When a die is thrown, we have \(S=\{1,2,3,4,5,6\}\).
Let \(E=\) event of getting a number greater than \(4=\{5,6\}\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{2}{6}=\frac{1}{3} .
\)
In a simultaneous throw of two dice, what is the probability of getting a total of 7?
We know that in a simultaneous throw of two dice, \(n\) (S) \(=6 \times 6=36\)
Let \(E=\) event of getting a total of 7
\(
\begin{aligned}
& =\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\} \\
& \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6} .
\end{aligned}
\)
What is the probability of getting a sum 9 from two throws of a dice?
In two throws of a die, \(n(\mathrm{~S})=(6 \times 6)=36\).
Let \(E=\) event of getting a sum 9
\(
\begin{aligned}
& =\{(3,6),(4,5),(5,4),(6,3)\} \\
& \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{4}{36}=\frac{1}{9}
\end{aligned}
\)
In a simultaneous throw of two dice, what is the probability of getting a doublet?
In a simultaneous throw of two dice, \(n(\mathrm{~S})=(6 \times 6)=36\).
Let \(E=\) event of getting a doublet \(=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}\).
In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11?
In a simultaneous throw of two dice, we have \(n\) (S) \(=(6 \times 6)=36\).
Let \(E=\) event of getting a total of 10 or 11 \(=\{(4,6),(5,5),(6,4),(5,6),(6,5)\}\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{5}{36}
\)
Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
In a simultaneous throw of two dice, we have \(n(\mathrm{~S})=\) \((6 \times 6)=36\)
Let \(E=\) event of getting two numbers whose product is even.
Then, \(E=\{(1,2),(1,4),(1,6),(2,1),(2,2),(2,3)\), \((2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2)\), \((4,3),(4,4),(4,5),(4,6),(5,2),(5,4),(5,6),(6,1)\), \((6,2),(6,3),(6,4),(6,5),(6,6)\}\)
\(\therefore \quad n(E)=27\)
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{27}{36}=\frac{3}{4}\).
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3?
Here, \(S=\{1,2,3,4, \ldots ., 19,20\}\)
Let \(E=\) event of getting a multiple of \(3=\{3,6,9,12,15,18\}\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{6}{20}=\frac{3}{10}
\)
Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?
Here, \(S=\{1,2,3,4, \ldots \ldots, 19,20\}\)
Let \(E=\) event of getting a multiple of 3 or \(5=\{3,6,9,12\), \(15,18,5,10,20\}\)
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{9}{20} .
\)
In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?
\(
P(\text { getting a prize })=\frac{10}{(10+25)}=\frac{10}{35}=\frac{2}{7}
\)
One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card?
Clearly, there are 52 cards, out of which there are 16 face cards.
\(
\therefore \quad P(\text { getting a face card })=\frac{16}{52}=\frac{4}{13} .
\)
A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is
Here, \(n(\mathrm{~S})=52\).
Let \(E=\) event of getting a queen of club or a king of heart. Then, \(\quad n(E)=2\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{2}{52}=\frac{1}{26} \text {. }
\)
One card is drawn from a pack of 52 cards. What is the probability that the card drawn is either a red card or a king?
Here, \(n(\mathrm{~S})=52\).
There are 26 red cards (including 2 kings) and there are 2 more kings.
Let \(E=\) event of getting a red card or a king.
Then, \(\quad n(E)=28\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{28}{52}=\frac{7}{13} .
\)
From a pack of 52 cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade?
Here, \(n(\mathrm{~S})=52\)
There are 13 spades (including one ten) and there are 3 more tens.
Let \(E=\) event of getting a ten or a spade.
Then, \(n(E)=(13+3)=16\).
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{16}{52}=\frac{4}{13}
\)
The probability that a card drawn from a pack of 52 cards will be a diamond or a king, is
Here, \(n(\mathrm{~S})=52\).
There are 13 cards of diamond (including one king) and there are 3 more kings.
Let \(E=\) event of getting a diamond or a king.
Then, \(\quad n(E)=(13+3)=16\)
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{16}{52}=\frac{4}{13} .
\)
From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
Let \(S\) be the sample space.
Then, \(n(S)={ }^{52} C_2=\frac{(52 \times 51)}{(2 \times 1)}=1326\).
Let \(E=\) event of getting 2 kings out of 4 .
\(
\begin{aligned}
& \therefore \quad n(E)={ }^4 C_2=\frac{(4 \times 3)}{(2 \times 1)}=6 . \\
& \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{6}{1326}=\frac{1}{221} .
\end{aligned}
\)
Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is
Let \(S\) be the sample space.
Then, \(n(S)={ }^{52} C_2=\frac{(52 \times 51)}{(2 \times 1)}=1326\).
Let \(E=\) event of getting 1 spade and 1 heart.
\(\therefore \quad n(E)=\) number of ways of choosing 1 spade out of 13 and 1 heart out of 13
\(
\begin{aligned}
& =\left({ }^{13} C_1 \times{ }^{13} C_1\right)=(13 \times 13)=169 . \\
& \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{169}{1326}=\frac{13}{102} .
\end{aligned}
\)
Two cards are drawn from a pack of 52 cards. The probability that either both are red or both are kings is
Clearly, \(n(S)={ }^{52} C_2=\frac{(52 \times 51)}{2}=1326\).
Let \(E_1=\) event of getting both red cards,
\(E_2=\) event of getting both kings.
Then, \(E_1 \cap E_2=\) event of getting 2 kings of red cards.
\(
\begin{aligned}
& \therefore n\left(E_1\right)={ }^{26} C_2=\frac{(26 \times 25)}{(2 \times 1)}=325 ; n\left(E_2\right)={ }^4 C_2=\frac{(4 \times 3)}{(2 \times 1)}=6 ; \\
& \quad n\left(E_1 \cap E_2\right)=2 C_2=1 \\
& \therefore \quad P\left(E_1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{325}{1326} ; P\left(E_2\right)=\frac{n\left(E_2\right)}{n(S)}=\frac{6}{1326} ; \\
& P\left(E_1 \cup E_2\right)=\frac{1}{1326} \\
& \therefore \quad P(\text { both red or both kings })=P\left(E_1 \cup E_2\right)=P\left(E_1\right)+ \\
& P\left(E_2\right)-P\left(E_1 \cap E_2\right) \\
& \quad=\left(\frac{325}{1326}+\frac{6}{1326}-\frac{1}{1326}\right)=\frac{330}{1326}=\frac{55}{221} .
\end{aligned}
\)
A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?
The total number of balls \(=(6+8)=14\). The number of white balls \(=8\).
\(
P(\text { drawing a white ball })=\frac{8}{14}=\frac{4}{7} \text {. }
\)
In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
Total number of balls \(=(8+7+6)=21\).
Let \(E=\) Event that the ball drawn is neither red nor green
\(=\) Event that the ball drawn is red.
\(
\begin{aligned}
& \therefore \quad n(E)=8 . \\
& \therefore \quad P(E)=\frac{8}{21} .
\end{aligned}
\)
A box contains 4 red, 5 green and 6 white balls. A ball is drawn at random from the box. What is the probability that the ball drawn is either red or green?
The total number of balls \(=(4+5+6)=15\). \(P(\) drawing a red ball or a green ball \()=P(\) red \()+P(\) green \()\) \(=\left(\frac{4}{15}+\frac{5}{15}\right)=\frac{9}{15}=\frac{3}{5}\)
A basket contains 4 red, 5 blue, and 3 green marbles. If 2 marbles are drawn at random from the basket, what is the probability that both are red? (S.B.I. P.O., 2010)
Total number of balls \(=(4+5+3)=12\) Let \(E\) be the event of drawing 2 red balls.
Then, \(n(E)={ }^4 C_2=\frac{4 \times 3}{2 \times 1}=6\).
Also, \(n(S)={ }^{12} C_2=\frac{12 \times 11}{2 \times 1}=66\).
\(\therefore P(E)=\frac{n(E)}{n(S)}=\frac{6}{66}=\frac{1}{11}\).
An urn contains 6 red, 4 blue, 2 green, and 3 yellow marbles. If two marbles are drawn at random from the urn, what is the probability that both are red? (S.B.I. P.O., 2010)
Total number of balls \(=(6+4+2+3)=15\). Let \(E\) be the event of drawing 2 red balls.
Then, \(n(E)={ }^6 C_2=\frac{6 \times 5}{2 \times 1}=15\).
Also, \(n(S)={ }^{15} C_2=\frac{15 \times 14}{2 \times 1}=105\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{15}{105}=\frac{1}{7}\).
A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow? (Bank P.O., 2009)
Total number of balls \(=(6+2+4+3)=15\).
Let \(E\) be the event of drawing 3 non-yellow balls.
Then, \(n(E)={ }^{12} C_3=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}=220\).
Also, \(n(S)={ }^{15} C_3=\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{220}{455}=\frac{44}{91}\).
An urn contains 6 red, 4 blue, 2 green, and 3 yellow marbles. If three marbles are picked up at random, what is the probability that 2 are blue and 1 is yellow? (S.B.I. P.O., 2010)
Total number of marbles \(=(6+4+2+3)=15\).
Let \(E\) be the event of drawing 2 blue and 1 yellow marble.
Then, \(n(E)=\left({ }^4 C_2 \times{ }^3 C_1\right)=\frac{4 \times 3}{2 \times 1} \times 3=18\).
Also, \(n(S)={ }^{15} C_3=\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{18}{455}\).
An urn contains 6 red, 4 blue, 2 green, and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red? (S.B.I. P.O., 2011)
Total number of marbles \(=(6+4+2+3)=15\).
Let \(E\) be the event of drawing 1 green, 2 blue and 1 red marble.
Then, \(n(E)=\left({ }^2 C_1 \times{ }^4 C_2 \times{ }^6 C_1\right)=2 \times \frac{4 \times 3}{2 \times 1} \times 6=72\).
And, \(n(S)={ }^{15} C_4=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=1365\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{72}{1365}=\frac{24}{455}\).
An urn contains 6 red, 4 blue, 2 green, and 3 yellow marbles. If two marbles are picked up at random, what is the probability that either both are green or both are yellow? (Bank P.O., 2010)
Total number of marbles \(=(6+4+2+3)=15\).
Let \(E\) be the event of drawing 2 marbles such that either both are green or both are yellow.
Then, \(n(E)=\left({ }^2 C_1+{ }^3 C_2\right)=\left(1+{ }^3 C_1\right)=(1+3)=4\). And, \(n(S)={ }^{15} C_2=\frac{15 \times 14}{2 \times 1}=105\)
\(\therefore \quad \mathrm{P}(\mathrm{E})=\frac{n(E)}{n(S)}=\frac{4}{105}\).
A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red and 2 are green? (Bank P.O., 2009)
Total number of balls \(=(6+2+4+3)=15\).
Let \(E\) be the event of drawing 4 balls such that 2 are red and 2 are green.
Then, \(n(E)=\left({ }^2 C_2 \times{ }^4 C_2\right)=\left(1 \times \frac{4 \times 3}{2 \times 1}\right)=6\).
And, \(n(S)={ }^{15} C_4=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=1365\)
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{6}{1365}=\frac{2}{455}\).
A basket contains 4 red, 5 blue, and 3 green marbles. If three marbles are picked up at random what is the probability that at least one is blue? (S.B.I. P.O., 2010)
Total number of marbles \(=(4+5+3)=12\)
Let \(E\) be the event of drawing 3 marbles such that none is blue.
Then, \(n(E)=\) number of ways of drawing 3 marbles out of \(7={ }^7 C_3=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35\).
And, \(n(S)={ }^{12} C_3=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}=220\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{35}{220}=\frac{7}{44}\).
Required probability \(=1-P(E)=\left(1-\frac{7}{44}\right)=\frac{37}{44}\)
An urn contains 6 red, 4 blue, 2 green, and 3 yellow marbles. If 4 marbles are picked up at random, what is the probability that at least one of them is blue? (S.B.I. P.O., 2010)
Total number of marbles \(=(6+4+2+3)=15\).
Let \(E\) be the event of drawing 4 marbles such that none is blue.
Then, \(n(E)=\) number of ways of drawing 4 marbles out of 11 non-blue
\(
\begin{aligned}
& \qquad \quad{ }^{11} C_4=\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}=330 \\
& \text { And, } n(S) \quad={ }^{15} C_4=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=1365 \\
& \therefore \quad P(E) \quad=P(E)=\frac{n(E)}{n(S)}=\frac{330}{1365}=\frac{22}{91} \\
& \therefore \quad \text { Required probability }=\left(1-\frac{22}{91}\right)=\frac{69}{91}
\end{aligned}
\)
A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue? (Bank P.O., 2009)
Total number of balls \(=(6+2+4+3)=15\).
Let \(E\) be the event of drawing 5 balls out of 9 non-blue balls
\(
\begin{aligned}
& ={ }^9 C_5={ }^9 C_{(9-5)}={ }^9 C_4=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}=126 . \\
& \text { And, } n(S)={ }^{15} C_5=\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}=3003 \\
& \therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{126}{3003}=\frac{6}{143}
\end{aligned}
\)
\(
\therefore \quad \text { Required probability }=\left(1-\frac{6}{143}\right)=\frac{137}{143} .
\)
An urn contains 2 red, 3 green, and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue. (Railways, 2006)
Total number of balls \(=(2+3+2)=7\).
Let \(E\) be the event of drawing 2 non-blue balls.
Then, \(n(E)={ }^5 C_2=\frac{5 \times 4}{2 \times 1}=10\).
And, \(n(S)={ }^7 C_2=\frac{7 \times 6}{2 \times 1}=21\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{10}{21}\)
A box contains 10 black and 10 white balls. What is the probability of drawing 2 balls of the same colour?
Total number of balls \(=(10+10)=20\).
Let \(E\) be the event of drawing 2 balls of the same colour. \(n(E)=\) number of ways of drawing 2 black balls or 2 white balls
\(
\begin{aligned}
n(E) & =\left({ }^{10} C_2+{ }^{10} C_2\right)=2 \times{ }^{10} C_2=2 \times \frac{10 \times 9}{2 \times 1}=90 \\
n(S) & =\text { number of ways of drawing } 2 \text { balls out of } 20 \\
& ={ }^{20} C_2=\frac{20 \times 19}{2 \times 1}=190 \\
\therefore \quad P(E) & =\frac{n(E)}{n(S)}=\frac{90}{190}=\frac{9}{19}
\end{aligned}
\)
A box contains 20 electric bulbs, out of which 4 are defective. Two balls are chosen at random from
this box. The probability that at least one of them is defective, is
\(
\begin{aligned}
& P \text { (none is defective) }=\frac{{ }^{16} C_2}{{ }^{20} C_2}=\left(\frac{16 \times 15}{2 \times 1} \times \frac{2 \times 1}{20 \times 19}\right)=\frac{12}{19} \\
& P\left(\text { at least } 1 \text { is defective) }=\left(1-\frac{12}{19}\right)=\frac{7}{19}\right.
\end{aligned}
\)
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that the selected students are 2 boys and 1 girl, is:
Let \(S\) be the sample space and let \(E\) be the event of selecting 2 boys and 1 girl.
Then, \(n(S) \quad=\) number of ways of selecting 3 students out of \(25={ }^{25} C_3=\frac{25 \times 24 \times 23}{3 \times 2 \times 1}=2300\).
And, \(n(E)=\left({ }^{15} C_2 \times{ }^{10} C_1\right)=\left(\frac{15 \times 14}{2 \times 1} \times 10\right)=1050\)
\(
\therefore \quad \mathrm{P}(\mathrm{E})=\frac{n(E)}{n(S)}=\frac{1050}{2300}=\frac{21}{46}
\)
Four persons are chosen at random from a group of 3 men, 2 women, and 4 children. The chance that exactly 2 of them are children, is
\(n(S)=\) number of ways of choosing 4 persons out of 9
\(
={ }^9 C_4=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}=126 \text {. }
\)
\(n(E)=\) Number of ways of choosing 2 children out of 4 and 2 persons out of \((3+2)\) persons
\(
\begin{aligned}
& n(E)=\left({ }^4 C_2 \times{ }^5 C_2\right)=\left(\frac{4 \times 3}{2 \times 1} \times \frac{5 \times 4}{2 \times 1}\right)=60 . \\
\therefore \quad & P(E)=\frac{n(E)}{n(S)}=\frac{60}{126}=\frac{10}{21} .
\end{aligned}
\)
Two dice are tossed. The probability that the total score is a prime number is
Clearly, \(n(S)=(6 \times 6)=36\)
Let \(E\) be the event that the sum is a prime number. Then, \(n(E)=\{(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2)\), \((3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)\}\)
\(
\therefore \quad n(E)=15 \text {. }
\)
\(
\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{15}{36}=\frac{5}{12} .
\)
In a class, \(30 \%\) of the students offered English, \(20 \%\) offered Hindi and \(10 \%\) offered both. If a student is selected at random, what is the probability that he has offered English or Hindi?
\(
P(E)=\frac{30}{100}=\frac{3}{10}, P(H)=\frac{20}{100}=\frac{1}{5} \text { and } P(E \cap H)=\frac{10}{100}=\frac{1}{10}
\)
\(
P(E \text { or } H)=P(E \cup H)
\)
\(
=P(E)+P(H)-P(E \cap H)=\left(\frac{3}{10}+\frac{1}{5}-\frac{1}{10}\right)=\frac{4}{10}=\frac{2}{5}
\)
A man and his wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is \(\frac{1}{7}\) and the probability of wife’s selection is \(\frac{1}{5}\). What is the probability that only one of them is selected?
Let \(E_1=\) Event that the husband is selected and \(E_2=\) Event that the wife is selected. Then,
\(
\begin{aligned}
& P\left(E_1\right)=\frac{1}{7} \text { and } P\left(E_2\right)=\frac{1}{5} . \\
\therefore & P\left(\bar{E}_1\right)=\left(1-\frac{1}{7}\right)=\frac{6}{7} \text { and } P\left(\bar{E}_2\right)=\left(1-\frac{1}{5}\right)=\frac{4}{5} . \\
\therefore & \text { Required probability }=P[(A \text { and not } B) \text { or }(B \text { and not } A)] \\
= & P\left[\left(E_1 \cap \bar{E}_2\right) \text { or }\left(E_2 \cap \bar{E}_1\right)\right] \\
= & P\left[\left(E_1 \cap \bar{E}_2\right)+P\left(E_2 \cap \bar{E}_1\right)\right] \\
= & P\left(E_1\right) \cdot P\left(\bar{E}_2\right)+P\left(E_2\right) \cdot P\left(\bar{E}_1\right)=\left(\frac{1}{7} \times \frac{4}{5}\right)+\left(\frac{1}{5} \times \frac{6}{7}\right)=\frac{10}{35}=\frac{2}{7} .
\end{aligned}
\)
A speaks truth in \(75 \%\) cases and B in \(80 \%\) of the cases. In what percentage of cases are they likely to contradict each other, in narrating the same incident?
Let \(E_1=\) Event that \(A[latex] speaks the truth and [latex]E_2=\) Event that \(B\) speaks the truth. Then, \(\begin{aligned} P\left(E_1\right) & =\frac{75}{100}=\frac{3}{4}, P\left(E_2\right)=\frac{80}{100}=\frac{4}{5}, P\left(\bar{E}_1\right)=\left(1-\frac{3}{4}\right) \\ & =\frac{1}{4}, P\left(\bar{E}_2\right)=\left(1-\frac{4}{5}\right)=\frac{1}{5}\end{aligned}\)
\(P(A\) and \(B\) contradict each other \()\)
\(=P[(A\) speaks the truth and \(B\) tells a lie) or ( \(A\) tells a lie and \(B\) speaks the truth)]
\(
\begin{aligned}
& =P\left[\left(E_1 \cap \bar{E}_2\right) \text { or }\left(\bar{E}_1 \cap E_2\right)\right]=P\left(E_1 \cap \bar{E}_2\right)+P\left(\bar{E}_1 \cap E_2\right) \\
& =P\left(E_1\right) \cdot P\left(\bar{E}_2\right)+P\left(\bar{E}_1\right) \cdot P\left(E_2\right) \\
& =\left(\frac{3}{4} \times \frac{1}{5}\right)+\left(\frac{1}{4} \times \frac{4}{5}\right)=\left(\frac{3}{20}+\frac{1}{5}\right)=\frac{7}{20} \\
& =\left(\frac{7}{20} \times 100\right) \%=35 \%
\end{aligned}
\)
A speaks truth in \(60 \%\) cases and B speaks truth in \(70 \%\) cases. The probability that they will say the same thing while describing a single event, is (Railways, 2006)
Let \(E_1=\) Event that \(\mathrm{A}\) speaks the truth and \(E_2=\) Event that B speaks the truth.
\(
\text { Then, } P\left(E_1\right)=\frac{60}{100}=\frac{3}{5}, P\left(E_2\right)=\frac{70}{100}=\frac{7}{10}, P\left(\bar{E}_1\right)
\)
\(
=\left(1-\frac{3}{5}\right)=\frac{2}{5}, P\left(\bar{E}_2\right)=\left(1-\frac{7}{10}\right)=\frac{3}{10} .
\)
\(P(A\) and \(B\) say the same thing)
\(=P[(A\) speaks the truth and \(B\) speaks the truth) or \((A\) tells a lie and \(B\) tells a lie)]
\(
\begin{aligned}
& =P\left[\left(E_1 \cap E_2\right) \text { or }\left(\bar{E}_1 \cap \bar{E}_2\right)\right]=P\left(E_1 \cap E_2\right)+P\left(\bar{E}_1 \cap \bar{E}_2\right) \\
& =P\left(E_1\right) \cdot P\left(E_2\right)+P\left(\bar{E}_1\right) \cdot P\left(\bar{E}_2\right) \\
& =\left(\frac{3}{5} \times \frac{7}{10}\right)+\left(\frac{2}{5} \times \frac{3}{10}\right)=\frac{27}{50}=0.54
\end{aligned}
\)
A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least 1 woman? (Bank P.O. ,2008)
Total number of persons \(=(3+2)=5\).
\(
\therefore \quad n(S)={ }^5 C_3={ }^5 C_2=\frac{5 \times 4}{2 \times 1}=10 .
\)
Let \(E\) be the event of selecting 3 members having at least 1 woman
Then, \(n(E)=n[(1\) woman and 2 men) or ( 2 women and 1 man)]
\(=n\) ( 1 woman and 2 men \()+n(2\) women and 1 man \()\)
\(=\left({ }^2 C_1 \times{ }^3 C_1\right)+\left({ }^2 C_2 \times{ }^3 C_1\right)=\left({ }^2 C_1 \times{ }^3 C_1\right)+\left(1 \times{ }^3 C_1\right)\)
\(=(2 \times 3)+(1 \times 3)=(6+3)=9\).
\(\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{9}{10}\).
A bag contains 3 blue, 2 green and 5 red balls. If four balls are picked at random, what is the probability that two are green and two are blue? [DMRC-Customer Relationship Assistant (CRA) Exam, 2016]
Number of blue balls \(=3\) balls
Number of green balls \(=2\) balls
Number of red balls \(=5\) balls
Total balls in the bag \(=3+2+5=10\)
Total possible outcomes \(=\) Selection of 4 balls out 10 balls \(={ }^{10} C_4=\frac{10 !}{4 \times(10-4) !}=\frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4}=210\)
Favorable outcomes \(=(\) selection of 2 green balls out of 2 balls) \(\times\) (selection of 2 balls out of 3 blue balls)
\(
\begin{aligned}
& ={ }^2 C_2 \times{ }^3 C_2 \\
& =1 \times 3=3 \\
& \therefore \text { Required probability }=\frac{\text { Favorable out comes }}{\text { Total possible outcomes }} \\
& \qquad=\frac{3}{210}=\frac{1}{70}
\end{aligned}
\)
Dev can hit a target 3 times in 6 shorts Pawan can hit the target 2 times in 6 shorts and Lakhan can hit the target 4 times in 4 shorts. What is the probability that at least 2 shorts hit the target [DMRC-Customer Relationship Assistant (CRA) Exam, 2016]
Probability of hitting the target:
Dev can hit target \(\Rightarrow \frac{3}{6}=\frac{1}{2}\), Lakhan can hit target \(=\frac{4}{4}=1\)
Pawan can hit target \(=\frac{2}{6}=\frac{1}{3}\)
Required probability that at least 2 shorts hit target
\(
\begin{aligned}
& =\frac{1}{2} \times \frac{2}{3}+\frac{1}{2} \times \frac{1}{3}+\frac{1}{2} \times \frac{1}{3} \\
& =\frac{1}{3}+\frac{1}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}
\end{aligned}
\)
A bag contains 10 mangoes out of which 4 are taken out together. If one of them is found to be good, the probability that other is also good is [DMRC-Train Operator (Station Controller) Exam, 2016]
Out of 10 mangoes, 4 mangoes are rotten
\(
\begin{aligned}
\therefore \text { Required probability }= & \frac{{ }^6 C_2}{{ }^{10} C_2}=\frac{\frac{6 !}{2 !(6-2) !}}{\frac{10 !}{2 !(10-2) !}}=\frac{\frac{6 !}{2 ! 4 !}}{\frac{10 !}{2 ! \times 8 !}} \\
= & \frac{\frac{6 \times 5}{1 \times 2}}{\frac{10 \times 9}{1 \times 2}}=\frac{6 \times 5}{10 \times 9}=\frac{1}{3}
\end{aligned}
\)
A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random. What is the probability that none of the balls drawn are yellow in colour? [IBPS-Bank PO/MT (Pre.) Exam, 2015]
Number of red balls \(=4\)
Number of yellow ball \(=5\)
Number of pink ball \(=6\)
Total number of balls \(=4+5+6=15\)
Total possible outcomes \(=\) selection of 2 balls out of 15
balls \(={ }^{15} C_2=\frac{15 !}{2 !(15-2) !}=\frac{15 !}{2 ! \times 13 !}=\frac{15 \times 14}{1 \times 2}=105\)
Total favourable outcomes \(=\) selection of 2 balls out of 4 orange and 6 pink balls.
\(
\begin{aligned}
& { }^{10} C_2=\frac{10 !}{2 !(10-2) !}=\frac{10 !}{2 ! 8 !}=\frac{10 \times 9}{1 \times 2}=45 \\
& \therefore \text { Required probability }=\frac{45}{105}=\frac{3}{7}
\end{aligned}
\)
A bag contains 6 red balls 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag. One after another what is the probability that the first ball is red and second ball is yellow. [IBPS-Bank PO (Pre.) Exam, 2015]
Number of red balls \(=6\)
Number of yellow ball \(=11\)
Number of pink balls \(=5\)
Total number of balls \(=6+11+5=22\)
Total possiblecutcomes \(=n(E)={ }^{22} C_2=\frac{22 !}{2 !(22-2) !}=\frac{22 !}{2 ! 20 !}\) \(=\frac{22 \times 21}{2 \times 1}=231\)
Number of favourable outcomes
\(
\begin{aligned}
\qquad=n(s) & ={ }^6 C_1 \times{ }^{11} C_1=6 \times 11=66 \\
\text { Required probability } & =\frac{n(E)}{n(S)}=\frac{66}{231}=\frac{2}{7}
\end{aligned}
\)
A bag contains 4 red balls, 6 blue balls and 8 pink balls. One ball is drawn at random and replace with 3 pink balls. A probability that the first ball drawn was either red or blue in colour and the second ball drawn was pink in colour? [CET-(Maharashtra (MBA) Exam, 2016]
\(
\begin{aligned}
& \text { Number of Red balls }=4 \\
& \text { Number of Blue balls }=6 \\
& \text { Number of Pink balls }=8 \\
& \text { Total number of balls }=4+6+8=18 \\
& \begin{aligned}
& \text { Required probability } \\
&=\frac{4}{18} \times \frac{11}{20}+\frac{6}{18} \times \frac{11}{20} \\
&=\frac{11}{20}\left[\left(\frac{4}{18}+\frac{6}{18}\right)\right] \\
&=\frac{11}{20} \times \frac{10}{18}=\frac{11}{36}
\end{aligned}
\end{aligned}
\)
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