Differentiation (Part-38) | S N De

Probability (Class 12): Solved Questions 27–36

Step-by-step solutions to ten questions on total probability, conditional probability, bag and urn transfer problems and games of chance. Each solution shows the formula used, the case table or working, and a boxed final answer. Useful for CBSE Class 12 board exam, JEE and other competitive exam preparation.

Q27

A person reports that during a test the probability of a microbe reacting to drug \(A\) is 0.62 and to drug \(B\) is 0.53. The probability that the microbe reacts to both drug \(A\) and \(B\) is 0.18 and that it does not react to any one of them is 0.13. Should the report of the test be questioned?

Solution

Let \(X\) and \(Y\) be the events that the microbe reacts to drug \(A\) and drug \(B\). Given \(P(X)=0.62,\ P(Y)=0.53,\ P(X\cap Y)=0.18\) and \(P(X^c\cap Y^c)=0.13\). We find \(P(X\cup Y)\) in two different ways.

Method 1: Addition theorem\[P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)=0.62+0.53-0.18=0.97.\]Method 2: Complementary event

"Reacts to at least one drug" is the complement of "reacts to neither":

\[P(X\cup Y)=1-P(X^c\cap Y^c)=1-0.13=0.87.\]

Answer: The same probability \(P(X\cup Y)\) comes out as \(0.97\) by one method and \(0.87\) by the other. The data are inconsistent, so yes, the report of the test should be questioned.

Q28

(i) Two cards are drawn at random from a well-shuffled pack of 52 cards and thrown away. Find the probability of drawing an ace from the remaining 50 cards.
(ii) A pair of fair dice is thrown. Find the probability that the sum is 10 or greater if a 5 appears on the first die.

Solution

(i) Ace from the remaining 50 cards

Let \(E_0, E_1, E_2\) be the events that 0, 1 and 2 aces are among the two discarded cards. Let \(A\) be the event of drawing an ace from the remaining 50 cards. The pack has 4 aces and 48 other cards, and \(\binom{52}{2}=1326\).

Event	\(P(E_i)\)	Aces left	\(P(A/E_i)\)
\(E_0\)	\(\frac{\binom{48}{2}}{1326}=\frac{1128}{1326}\)	4	\(\frac{4}{50}\)
\(E_1\)	\(\frac{4\times48}{1326}=\frac{192}{1326}\)	3	\(\frac{3}{50}\)
\(E_2\)	\(\frac{\binom{4}{2}}{1326}=\frac{6}{1326}\)	2	\(\frac{2}{50}\)

\[P(A)=\frac{1128\times4+192\times3+6\times2}{1326\times50}=\frac{5100}{66300}=\frac1{13}.\]

Answer (i): \(\dfrac{1}{13}\).

Shortcut: by symmetry, the 50 remaining cards are as likely to contain an ace as a random 50 of the 52, so the answer equals \(\frac{4}{52}=\frac1{13}\).

(ii) Sum 10 or greater, given a 5 on the first die

Let \(F\) be the event "5 on the first die", so \(F=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}\) and \(n(F)=6\). Let \(E\) be the event "sum is 10 or greater". Among the outcomes in \(F\), the sum is at least 10 for \((5,5)\) and \((5,6)\), so \(n(E\cap F)=2\).

\[P(E/F)=\frac{n(E\cap F)}{n(F)}=\frac26=\frac13.\]

Answer (ii): \(\dfrac{1}{3}\).

Q29

(i) Three lots contain respectively 4%, 5% and 10% defective articles. One article is chosen at random from each lot. Find the probability of getting (a) exactly one defective, (b) at least one defective article among the three articles drawn.
(ii) An anti-aircraft gun can take a maximum of four shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1 respectively. What is the probability that the gun hits the plane?

Solution

(i) Three lots

The probabilities of a defective article are \(0.04,\ 0.05,\ 0.10\), and of a good article \(0.96,\ 0.95,\ 0.90\). The three draws are independent.

(a) Exactly one defective:

\[\begin{aligned}P&=0.04\times0.95\times0.90+0.96\times0.05\times0.90+0.96\times0.95\times0.10\\&=0.0342+0.0432+0.0912=0.1686.\end{aligned}\]

(b) At least one defective:

\[P=1-P(\text{none defective})=1-0.96\times0.95\times0.90=1-0.8208=0.1792.\]

Answer (i): (a) \(0.1686\), (b) \(0.1792\).

(ii) Gun hits the plane

The gun hits the plane if at least one of the four shots hits. The shots miss with probabilities \(0.6,\ 0.7,\ 0.8,\ 0.9\), and the shots are independent.

\[P(\text{hit})=1-0.6\times0.7\times0.8\times0.9=1-0.3024=0.6976.\]

Answer (ii): \(0.6976\).

Q30

An article manufactured by a company consists of two parts \(A\) and \(B\). In the process of manufacture of part \(A\), 9 out of 100 are likely to be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part \(B\). Calculate the probability that the assembled parts will not be defective.

Solution

\(P(A\text{ defective})=\frac{9}{100}\), so \(P(A\text{ not defective})=\frac{91}{100}\). \(P(B\text{ defective})=\frac{5}{100}\), so \(P(B\text{ not defective})=\frac{95}{100}\).

The assembled article is not defective when both parts are not defective. The two parts are manufactured independently, so

\[P=\frac{91}{100}\times\frac{95}{100}=\frac{8645}{10000}=0.8645.\]

Answer: \(0.8645\).

Q31CBSE '07

Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys, and 1 girl and 3 boys. One child is selected at random from each group. Find the chance that the selected group consists of 1 girl and 2 boys.

Solution

The probabilities for each group are

Group	\(P(\text{girl})\)	\(P(\text{boy})\)
1st (3 girls, 1 boy)	\(\frac34\)	\(\frac14\)
2nd (2 girls, 2 boys)	\(\frac24=\frac12\)	\(\frac12\)
3rd (1 girl, 3 boys)	\(\frac14\)	\(\frac34\)

One girl and two boys can be selected in three mutually exclusive ways:

Girl from	Probability
1st group	\(\frac34\times\frac12\times\frac34=\frac{9}{32}\)
2nd group	\(\frac14\times\frac12\times\frac34=\frac{3}{32}\)
3rd group	\(\frac14\times\frac12\times\frac14=\frac{1}{32}\)
Total	\(\frac{13}{32}\)

Answer: \(\dfrac{13}{32}\).

Q32

(i) A six faced die is so biased that it is twice as likely to show an even number as an odd. It is thrown twice. What is the probability that the sum of the numbers is even?
(ii) Three faces of a fair die are yellow, two faces red and one blue. The die is tossed three times. Find the probability that the colours yellow, red and blue appear in the 1st, 2nd and the 3rd tosses respectively.

Solution

(i) Biased die

Let \(P(\text{odd})=p\). Then \(P(\text{even})=2p\), and \(p+2p=1\) gives \(p=\frac13\). So \(P(\text{odd})=\frac13\) and \(P(\text{even})=\frac23\).

The sum of two numbers is even when both are even or both are odd:

\[P=\frac23\times\frac23+\frac13\times\frac13=\frac49+\frac19=\frac59.\]

Answer (i): \(\dfrac{5}{9}\).

(ii) Colours on a fair die

For a fair die, \(P(\text{yellow})=\frac36,\ P(\text{red})=\frac26,\ P(\text{blue})=\frac16\). The tosses are independent, so

\[P=\frac36\times\frac26\times\frac16=\frac12\times\frac13\times\frac16=\frac{1}{36}.\]

Answer (ii): \(\dfrac{1}{36}\).

Q33CBSE '03 (part i)

(i) \(A\) and \(B\) play alternately with an unbiased die. He who first throws a "six" wins. If \(A\) begins, show that his chance of winning is \(\frac{6}{11}\).
(ii) \(A, B\) and \(C\) in order toss an unbiased coin. The first who throws "head" wins. Find the respective probabilities of winning.

Solution

(i) Chance of A winning

On each throw, \(P(\text{six})=\frac16\) and \(P(\text{not six})=\frac56\). \(A\) wins on the 1st, 3rd, 5th, \(\dots\) throws. He wins on the 3rd throw if the first two throws are not sixes, and so on. So

\[P(A\text{ wins})=\frac16+\frac56\cdot\frac56\cdot\frac16+\left(\frac56\right)^4\frac16+\cdots\]

This is an infinite geometric series with first term \(a=\frac16\) and common ratio \(r=\frac{25}{36}\). Its sum is \(\dfrac{a}{1-r}\):

\[P(A\text{ wins})=\frac{1/6}{1-25/36}=\frac{1/6}{11/36}=\frac{6}{11}.\]

Hence shown: \(A\)'s chance of winning is \(\dfrac{6}{11}\).

(ii) Three players tossing a coin

On each toss, \(P(H)=P(T)=\frac12\). The order of tossing is \(A, B, C, A, B, C, \dots\) The first head wins.

\[P(A\text{ wins})=\frac12+\left(\frac12\right)^3\cdot\frac12+\left(\frac12\right)^6\cdot\frac12+\cdots=\frac{1/2}{1-1/8}=\frac{1/2}{7/8}=\frac47.\]

\(B\) wins on the 2nd, 5th, 8th, \(\dots\) tosses, i.e. after \(A\) has failed once:

\[P(B\text{ wins})=\frac12\cdot\frac12+\left(\frac12\right)^4\cdot\frac12+\cdots=\frac{1/4}{1-1/8}=\frac27.\]

\(C\) wins on the 3rd, 6th, 9th, \(\dots\) tosses:

\[P(C\text{ wins})=\left(\frac12\right)^3+\left(\frac12\right)^6+\cdots=\frac{1/8}{1-1/8}=\frac17.\]

Answer (ii): \(P(A)=\dfrac47,\ P(B)=\dfrac27,\ P(C)=\dfrac17\). Check: \(\frac47+\frac27+\frac17=1\).

Q34CBSE '02

A bag contains 5 red and 4 yellow balls. A ball is drawn at random from the bag and put into another bag which contains 3 red and 6 yellow balls. A ball is drawn randomly from the second bag. What is the probability that it is yellow?

Solution

Let \(E_1\) be the event that a red ball is transferred and \(E_2\) that a yellow ball is transferred. Then \(P(E_1)=\frac59\) and \(P(E_2)=\frac49\). Let \(Y\) be the event that the ball drawn from the second bag is yellow.

Transferred	Second bag then has	\(P(Y/E_i)\)
Red	4 red, 6 yellow (10 balls)	\(\frac{6}{10}\)
Yellow	3 red, 7 yellow (10 balls)	\(\frac{7}{10}\)

By the theorem of total probability,

\[P(Y)=\frac59\times\frac6{10}+\frac49\times\frac7{10}=\frac{30}{90}+\frac{28}{90}=\frac{58}{90}=\frac{29}{45}.\]

Answer: \(\dfrac{29}{45}\).

Q35

(i) Each of the two identical bags contains 5 white and 5 red balls. One ball is transferred at random without notice from the second bag to the first bag. Then one ball is drawn from the first bag. Find the probability that the ball drawn is a red ball.
(ii) An urn \(A\) contains 3 white and 5 red marbles. Another urn \(B\) contains 5 white and 3 red marbles. Two marbles are transferred from \(A\) to \(B\) and then one is drawn from \(B\). What is the probability that it is red?

Solution

(i) Two bags with 5 white and 5 red balls

The ball moved from the second bag is red or white with probability \(\frac12\) each. After the transfer, the first bag has 11 balls.

Transferred ball	First bag then has	\(P(\text{red drawn})\)
Red (prob. \(\frac12\))	6 red, 5 white	\(\frac{6}{11}\)
White (prob. \(\frac12\))	5 red, 6 white	\(\frac{5}{11}\)

\[P(\text{red})=\frac12\times\frac6{11}+\frac12\times\frac5{11}=\frac12\times\frac{11}{11}=\frac12.\]

Answer (i): \(\dfrac{1}{2}\).

(ii) Two urns

Urn \(A\) has 8 marbles (3 white, 5 red), so two marbles can be taken in \(\binom82=28\) ways. After the transfer, urn \(B\) has \(8+2=10\) marbles. Initially urn \(B\) has 5 white and 3 red.

Marbles transferred	Probability	Red in urn B	\(P(\text{red drawn})\)
2 red	\(\frac{\binom52}{28}=\frac{10}{28}\)	5	\(\frac{5}{10}\)
1 red, 1 white	\(\frac{5\times3}{28}=\frac{15}{28}\)	4	\(\frac{4}{10}\)
2 white	\(\frac{\binom32}{28}=\frac{3}{28}\)	3	\(\frac{3}{10}\)

\[P(\text{red})=\frac{10}{28}\cdot\frac5{10}+\frac{15}{28}\cdot\frac4{10}+\frac3{28}\cdot\frac3{10}=\frac{50+60+9}{280}=\frac{119}{280}=\frac{17}{40}.\]

Answer (ii): \(\dfrac{17}{40}\).

Q36

There are three bags each containing 5 red and 5 black balls. One ball is transferred blindly from the first to the second bag, and then one ball is transferred blindly from the second to the third. A ball is now drawn from the third bag. Find the probability that the drawn ball is red.

Solution

Track the colour of the first transferred ball (from bag 1 to bag 2), then the colour of the second transferred ball (from bag 2 to bag 3). Bag 2 has 11 balls after the first transfer, and bag 3 has 11 balls after the second.

1st transfer	Bag 2 then has	2nd transfer	Bag 3 then has	P(this path and red drawn)
Red (\(\frac12\))	6 R, 5 B	Red (\(\frac6{11}\))	6 R, 5 B	\(\frac12\cdot\frac6{11}\cdot\frac6{11}=\frac{36}{242}\)
Red (\(\frac12\))	6 R, 5 B	Black (\(\frac5{11}\))	5 R, 6 B	\(\frac12\cdot\frac5{11}\cdot\frac5{11}=\frac{25}{242}\)
Black (\(\frac12\))	5 R, 6 B	Red (\(\frac5{11}\))	6 R, 5 B	\(\frac12\cdot\frac5{11}\cdot\frac6{11}=\frac{30}{242}\)
Black (\(\frac12\))	5 R, 6 B	Black (\(\frac6{11}\))	5 R, 6 B	\(\frac12\cdot\frac6{11}\cdot\frac5{11}=\frac{30}{242}\)

In the last column the chance of drawing red from bag 3 is \(\frac6{11}\) (when it has 6 R) or \(\frac5{11}\) (when it has 5 R). Adding the four paths, the probability of drawing red is

\[\frac12\left[\frac{6}{11}\cdot\frac{6}{11}+\frac{5}{11}\cdot\frac{5}{11}+\frac{5}{11}\cdot\frac{6}{11}+\frac{6}{11}\cdot\frac{5}{11}\right]=\frac12\times\frac{36+25+30+30}{121}=\frac12\times\frac{121}{121}=\frac12.\]

Answer: \(\dfrac{1}{2}\).

Why \(\tfrac12\)? All bags start with equal numbers of red and black balls, so by symmetry red and black are equally likely at every stage.

Quick recap: If \(E_1,\dots,E_n\) partition the sample space, \(P(A)=\sum P(E_i)\,P(A/E_i)\). For "first to succeed wins" games, add the infinite geometric series \(\dfrac{a}{1-r}\).