Differentiation (Part-38) | S N De

Probability (Class 12): Solved Questions 8–26

Step-by-step solutions to nineteen questions on conditional probability, independent events, odds, the addition theorem and the theorem of total probability. Each solution is written with the formula used, the working and a boxed final answer. Useful for CBSE Class 12 board exam, JEE and other competitive exam preparation.

Q8

If \(A\) and \(B\) are two events and \(P(B)\ne1\), prove that \(P(A/B^c)=\dfrac{P(A)-P(A\cap B)}{1-P(B)}\). Hence, deduce that \(P(A\cap B)>P(A)+P(B)-1\).

Solution

Part 1: Proof

By the definition of conditional probability, and since \(P(B^c)=1-P(B)\ne0\),

\[P(A/B^c)=\frac{P(A\cap B^c)}{P(B^c)}=\frac{P(A\cap B^c)}{1-P(B)}.\]

The event \(A\) is the union of the two mutually exclusive events \(A\cap B\) and \(A\cap B^c\). Hence \(P(A)=P(A\cap B)+P(A\cap B^c)\), i.e. \(P(A\cap B^c)=P(A)-P(A\cap B)\). Therefore

\[P(A/B^c)=\frac{P(A)-P(A\cap B)}{1-P(B)}.\]Part 2: Deduction

Every probability is at most 1, so \(P(A/B^c)\le1\):

\[\frac{P(A)-P(A\cap B)}{1-P(B)}\le1.\]

Since \(1-P(B)>0\), multiplying both sides by it gives \(P(A)-P(A\cap B)\le1-P(B)\), that is

\[P(A\cap B)\ge P(A)+P(B)-1.\]The inequality is strict, \(P(A\cap B)>P(A)+P(B)-1\), whenever \(P(A/B^c)<1\).

Note: equality occurs only in the extreme case \(P(A/B^c)=1\), i.e. when \(A\cup B\) is certain. In all other cases the strict inequality asked in the question holds.

Q9

If \(A\) and \(B\) are two mutually exclusive events and \(P(A\cup B)\ne0\), then show that \(P[A/(A\cup B)]=\dfrac{P(A)}{P(A)+P(B)}\).

Solution

By the definition of conditional probability, since \(P(A\cup B)\ne0\),

\[P[A/(A\cup B)]=\frac{P[A\cap(A\cup B)]}{P(A\cup B)}.\]

Since \(A\subset A\cup B\), we have \(A\cap(A\cup B)=A\). Since \(A\) and \(B\) are mutually exclusive, \(P(A\cup B)=P(A)+P(B)\). Hence

\[P[A/(A\cup B)]=\frac{P(A)}{P(A)+P(B)}.\]Hence shown.

Q10CBSE '05

\(X\) is taking up subjects: mathematics, physics and chemistry in the examination. His probabilities of getting grade \(A\) in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets grade \(A\) in (i) all subjects, (ii) no subject, (iii) two subjects.

Solution

Let \(M, P, C\) be the events of getting grade \(A\) in mathematics, physics and chemistry. Assuming the subjects are independent, \(P(M)=0.2,\ P(P)=0.3,\ P(C)=0.5\), and so \(P(M^c)=0.8,\ P(P^c)=0.7,\ P(C^c)=0.5\).

(i) Grade A in all subjects\[P(M\cap P\cap C)=0.2\times0.3\times0.5=0.03.\](ii) Grade A in no subject\[P(M^c\cap P^c\cap C^c)=0.8\times0.7\times0.5=0.28.\](iii) Grade A in exactly two subjects

Case	Probability
\(M,P\) yes, \(C\) no	\(0.2\times0.3\times0.5=0.03\)
\(M,C\) yes, \(P\) no	\(0.2\times0.7\times0.5=0.07\)
\(P,C\) yes, \(M\) no	\(0.8\times0.3\times0.5=0.12\)
Total	\(0.22\)

Answer: (i) 0.03, (ii) 0.28, (iii) 0.22.

Q11

From 100 tickets numbered \(1, 2, 3, \dots, 100\), four are drawn at random. What is the probability that 2 of them will bear numbers from 1 to 40 and the other 2 will bear any number from 41 to 100?

Solution

Four tickets can be drawn from 100 in \(\binom{100}{4}\) ways. There are 40 tickets numbered 1 to 40 and 60 tickets numbered 41 to 100. Favourable draws: 2 tickets from the first 40 and 2 tickets from the other 60.

\[\binom{40}{2}=780,\qquad \binom{60}{2}=1770,\qquad \binom{100}{4}=3{,}921{,}225.\]

\[P=\frac{\binom{40}{2}\binom{60}{2}}{\binom{100}{4}}=\frac{780\times1770}{3921225}=\frac{18408}{52283}\approx0.352.\]

Q12

Let \(A\) and \(B\) be independent events with \(P(A\cup B)=0.58\) and \(P(A\cap B)=0.12\). Determine the possible values of \(P(A)\).

Solution

Let \(P(A)=p\) and \(P(B)=q\). Since \(A\) and \(B\) are independent, \(P(A\cap B)=pq=0.12\).

From \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\):

\[p+q=0.58+0.12=0.70.\]

So \(p\) and \(q\) are the roots of \(t^2-0.7t+0.12=0\):

\[t=\frac{0.7\pm\sqrt{0.49-0.48}}{2}=\frac{0.7\pm0.1}{2}=0.4\ \text{or}\ 0.3.\]

Answer: \(P(A)=0.4\) (when \(P(B)=0.3\)) or \(P(A)=0.3\) (when \(P(B)=0.4\)).

Q13

Events \(A, B, C\) are such that \(P(A)=0.3,\ P(B)=0.4,\ P(C)=0.8,\ P(A\cap B)=0.08,\ P(A\cap C)=0.28\) and \(P(A\cap B\cap C)=0.09\). If \(P(A\cup B\cup C)\ge0.75\), show that \(0.23\le P(B\cap C)\le0.48\).

Solution

Step 1: Use the addition theorem\[\begin{aligned}P(A\cup B\cup C)&=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\\&=0.3+0.4+0.8-0.08-P(B\cap C)-0.28+0.09\\&=1.23-P(B\cap C).\end{aligned}\]Step 2: Upper limit

Given \(P(A\cup B\cup C)\ge0.75\), so \(1.23-P(B\cap C)\ge0.75\), which gives \(P(B\cap C)\le0.48\).

Step 3: Lower limit

Every probability is at most 1, so \(P(A\cup B\cup C)\le1\). Then \(1.23-P(B\cap C)\le1\), which gives \(P(B\cap C)\ge0.23\).

Combining Steps 2 and 3: \[0.23\le P(B\cap C)\le0.48.\]Hence shown.

Note: the data as printed give \(P(A\cap B\cap C)=0.09\), which is larger than \(P(A\cap B)=0.08\), though \(A\cap B\cap C\subset A\cap B\). This looks like a misprint in the book. The required result follows from the formula with the given numbers, as done above.

Q14

\(A, B, C\) and \(D\) are four mutually exclusive and exhaustive events. If the odds against the events \(B, C\) and \(D\) are \(7:2,\ 7:5\) and \(13:5\) respectively, find the odds in favour of the event \(A\).

Solution

If the odds against an event are \(m:n\), its probability is \(\dfrac{n}{m+n}\). So

\[P(B)=\frac{2}{9},\qquad P(C)=\frac{5}{12},\qquad P(D)=\frac{5}{18}.\]

Since the events are mutually exclusive and exhaustive, \(P(A)+P(B)+P(C)+P(D)=1\). With denominator 36:

\[P(B)+P(C)+P(D)=\frac{8}{36}+\frac{15}{36}+\frac{10}{36}=\frac{33}{36}=\frac{11}{12}.\]

Hence \(P(A)=1-\dfrac{11}{12}=\dfrac{1}{12}\) and \(P(A^c)=\dfrac{11}{12}\).

Odds in favour of \(A\) \(=P(A):P(A^c)=\dfrac1{12}:\dfrac{11}{12}=\) 1 : 11.

Q15CBSE '99C (part ii)NCERT (part iii)

(i) A problem on mathematics is given to three students \(A, B, C\) whose probabilities of solving it are \(\frac13,\frac25\) and \(\frac34\) respectively. Find the probability that the problem is solved.
(ii) A problem in mathematics is given to three students whose chances of solving it correctly are \(\frac12,\frac13\) and \(\frac14\) respectively. What is the probability that only one of them solves it correctly?
(iii) An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Solution

(i) Problem is solved

The problem is solved if at least one student solves it. The students work independently, so

\[P(\text{solved})=1-P(\text{none solves})=1-\frac23\times\frac35\times\frac14=1-\frac1{10}=\frac{9}{10}.\]

Answer (i): \(\dfrac{9}{10}\).

(ii) Only one solves it

Probabilities of failing are \(\frac12,\frac23,\frac34\). Exactly one student succeeds in three ways:

\[\begin{aligned}P&=\frac12\cdot\frac23\cdot\frac34+\frac12\cdot\frac13\cdot\frac34+\frac12\cdot\frac23\cdot\frac14\\&=\frac{6}{24}+\frac{3}{24}+\frac{2}{24}=\frac{11}{24}.\end{aligned}\]

Answer (ii): \(\dfrac{11}{24}\).

(iii) Easy question given it is multiple choice

Total questions \(=300+200+500+400=1400\). Let \(E\) be the event "easy question" and \(M\) the event "multiple choice question". Then \(n(M)=500+400=900\) and \(n(E\cap M)=500\).

\[P(E/M)=\frac{P(E\cap M)}{P(M)}=\frac{500/1400}{900/1400}=\frac{5}{9}.\]

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

Q16

A candidate is selected for interview for three posts. For the first post there are 3 candidates, for the second there are 4 and for the third there are 2. What are the chances of his getting at least one post?

Solution

He is one of the candidates for each post, and each post is filled independently with equal chance for every candidate. So his chances of getting the three posts are \(\frac13,\frac14,\frac12\), and his chances of missing them are \(\frac23,\frac34,\frac12\).

\[P(\text{at least one post})=1-P(\text{no post})=1-\frac23\times\frac34\times\frac12=1-\frac14.\]

Answer: \(\dfrac{3}{4}\).

Q17

(i) A bag contains 2 red and 3 white balls and another bag contains 1 red and 2 white balls. If a bag is chosen at random and a ball is drawn from it, what is the probability that the ball is white?
(ii) There are four boxes of one dozen eggs each. They contain 2, 3, 1, 0 spoiled eggs respectively. One box is selected at random and then an egg is taken at random. What is the probability that the egg is spoiled?

Solution

(i) Probability the ball is white

Let \(E_1, E_2\) be the events of choosing bag I and bag II, so \(P(E_1)=P(E_2)=\frac12\). Let \(W\) be the event of drawing a white ball. Then \(P(W/E_1)=\frac35\) and \(P(W/E_2)=\frac23\). By the theorem of total probability,

\[P(W)=\frac12\times\frac35+\frac12\times\frac23=\frac{3}{10}+\frac13=\frac{19}{30}.\]

Answer (i): \(\dfrac{19}{30}\).

(ii) Probability the egg is spoiled

Let \(B_1,\dots,B_4\) be the events of choosing each box, with \(P(B_i)=\frac14\). Let \(S\) be the event that the egg is spoiled. Each box has 12 eggs, so \(P(S/B_1)=\frac2{12},\ P(S/B_2)=\frac3{12},\ P(S/B_3)=\frac1{12},\ P(S/B_4)=0\).

\[P(S)=\frac14\left(\frac{2}{12}+\frac{3}{12}+\frac{1}{12}+0\right)=\frac14\times\frac{6}{12}=\frac18.\]

Answer (ii): \(\dfrac{1}{8}\).

Q18

Two consecutive drawings of a digit are made at random from the ten digits \(0, 1, 2, \dots, 9\). Find the probability that the product of the chosen digits is zero, given that the first drawn digit is replaced before the second drawing.

Solution

Because the first digit is replaced, there are \(10\times10=100\) equally likely outcomes. The product is zero when at least one of the two digits is 0.

The product is non-zero only when neither digit is 0, which happens in \(9\times9=81\) outcomes. Hence

\[P(\text{product}=0)=1-\frac{81}{100}=\frac{19}{100}.\]

Check: outcomes with at least one zero \(=10+10-1=19\).

Answer: \(\dfrac{19}{100}\).

Q19

Two digits are chosen at random from the digits \(1, 2, 3, \dots, 9\). If the sum of the chosen digits is odd, find the probability that one of the chosen digits is 6.

Solution

Let \(E\) be the event "the sum is odd" and \(F\) the event "one of the digits is 6". The sum of two digits is odd only if one is even and the other is odd.

Among \(1,\dots,9\) there are 4 even digits (\(2,4,6,8\)) and 5 odd digits (\(1,3,5,7,9\)).

\[n(E)=4\times5=20\ \text{pairs}.\]

For \(E\cap F\): one digit is 6 and the other is any of the 5 odd digits, so \(n(E\cap F)=1\times5=5\).

\[P(F/E)=\frac{n(E\cap F)}{n(E)}=\frac{5}{20}=\frac14.\]

Answer: \(\dfrac{1}{4}\).

Q20

The chance of an accident in three factories \(A, B, C\) in a year is 5 in 25, 6 in 36 and 8 in 64 respectively. Find the chances that an accident may happen in (i) at least one of them, (ii) all of them.

Solution

The probabilities of an accident are \(P(A)=\frac{5}{25}=\frac15,\ P(B)=\frac{6}{36}=\frac16,\ P(C)=\frac{8}{64}=\frac18\). Assume the factories are independent. Then the probabilities of no accident are \(\frac45,\ \frac56,\ \frac78\).

(i) At least one accident\[P=1-\frac45\times\frac56\times\frac78=1-\frac{140}{240}=1-\frac7{12}=\frac{5}{12}.\](ii) Accidents in all three\[P=\frac15\times\frac16\times\frac18=\frac{1}{240}.\]

Answer: (i) \(\dfrac{5}{12}\), (ii) \(\dfrac{1}{240}\).

Q21

The probability that a candidate passes in physics is 70% and that he passes in chemistry is 40%. What is the probability that the candidate passes in any one of the two subjects?

Solution

Let \(P\) and \(C\) be the events of passing in physics and chemistry. Assume they are independent: \(P(P)=0.7,\ P(C)=0.4\), so \(P(P^c)=0.3,\ P(C^c)=0.6\).

"Passes in any one of the two subjects" means he passes in at least one of them:

\[P(P\cup C)=1-P(P^c)P(C^c)=1-0.3\times0.6=1-0.18=0.82.\]

Check: \(P(P)+P(C)-P(P)P(C)=0.7+0.4-0.28=0.82\).

Answer: \(0.82\) (82%).

Note: if the question meant "exactly one" subject, the answer would be \(0.7\times0.6+0.3\times0.4=0.54\).

Q22CBSE '07 (part ii)

(i) A box contains 7 white and 5 black balls. If 3 balls are drawn simultaneously at random, what is the probability that they are not all of the same colour? Calculate the probability of the same event for the case where the balls are drawn in succession with replacement between drawings.
(ii) An urn contains 4 red and 7 black balls. Two balls are drawn at random with replacement. Find the probability of getting (a) 2 red balls, (b) 2 black balls, (c) one red ball and one black ball.

Solution

(i) Drawn simultaneously

Total ways \(=\binom{12}{3}=220\). All of the same colour: all white \(\binom73=35\) or all black \(\binom53=10\).

\[P(\text{not all same colour})=1-\frac{35+10}{220}=1-\frac{9}{44}=\frac{35}{44}.\](i) Drawn with replacement

Here \(P(\text{white})=\frac7{12}\) and \(P(\text{black})=\frac5{12}\) on every draw.

\[P(\text{all same})=\left(\frac7{12}\right)^3+\left(\frac5{12}\right)^3=\frac{343+125}{1728}=\frac{13}{48}.\]\[P(\text{not all same colour})=1-\frac{13}{48}=\frac{35}{48}.\]

Answer (i): \(\dfrac{35}{44}\) (simultaneously) and \(\dfrac{35}{48}\) (with replacement).

(ii) Urn with 4 red and 7 black balls

On each draw, \(P(R)=\frac{4}{11}\) and \(P(B)=\frac{7}{11}\), and the draws are independent because of replacement.

• (a) \(P(2\text{ red})=\dfrac4{11}\times\dfrac4{11}=\dfrac{16}{121}\)
• (b) \(P(2\text{ black})=\dfrac7{11}\times\dfrac7{11}=\dfrac{49}{121}\)
• (c) \(P(\text{one red, one black})=2\times\dfrac4{11}\times\dfrac7{11}=\dfrac{56}{121}\)

Answer (ii): (a) \(\dfrac{16}{121}\), (b) \(\dfrac{49}{121}\), (c) \(\dfrac{56}{121}\). Check: \(16+49+56=121\).

Q23

There are three persons aged 50, 60 and 70 years. The probability to live 10 years more is 0.8 for a 50 years old, 0.5 for a 60 years old and 0.2 for a 70 years old person. Find the probability that at least two of the three persons will remain alive 10 years hence.

Solution

Let the three persons be \(P_1, P_2, P_3\) (aged 50, 60, 70). The probabilities of being alive are \(0.8,\ 0.5,\ 0.2\), and of not being alive are \(0.2,\ 0.5,\ 0.8\). The persons are assumed independent.

"At least two alive" means exactly two alive or all three alive.

Case	Probability
\(P_1,P_2\) alive, \(P_3\) not	\(0.8\times0.5\times0.8=0.32\)
\(P_1,P_3\) alive, \(P_2\) not	\(0.8\times0.5\times0.2=0.08\)
\(P_2,P_3\) alive, \(P_1\) not	\(0.2\times0.5\times0.2=0.02\)
All three alive	\(0.8\times0.5\times0.2=0.08\)
Total	\(0.50\)

Answer: \(0.5\).

Q24

The probabilities for \(A, B\) and \(C\) hitting a target are \(\frac13,\frac15\) and \(\frac14\) respectively. If they try together, find the probability of exactly one shot hitting the target.

Solution

The probabilities of missing are \(\frac23,\frac45,\frac34\). Exactly one shot hits in three mutually exclusive ways, and the shooters are independent:

\[\begin{aligned}P&=\frac13\cdot\frac45\cdot\frac34+\frac23\cdot\frac15\cdot\frac34+\frac23\cdot\frac45\cdot\frac14\\&=\frac{12}{60}+\frac{6}{60}+\frac{8}{60}=\frac{26}{60}=\frac{13}{30}.\end{aligned}\]

Answer: \(\dfrac{13}{30}\).

Q25

The odds that a book will be favourably reviewed by three independent critics are 5 to 2, 4 to 3 and 3 to 4 respectively. What is the probability that the majority of the three will be favourable?

Solution

Odds in favour \(m:n\) give probability \(\dfrac{m}{m+n}\). So the probabilities of a favourable review are \(\frac57,\frac47,\frac37\), and of an unfavourable review are \(\frac27,\frac37,\frac47\).

A majority is favourable if at least two critics are favourable.

Case	Probability
All three favourable	\(\frac{5\cdot4\cdot3}{343}=\frac{60}{343}\)
1st, 2nd favourable, 3rd not	\(\frac{5\cdot4\cdot4}{343}=\frac{80}{343}\)
1st, 3rd favourable, 2nd not	\(\frac{5\cdot3\cdot3}{343}=\frac{45}{343}\)
2nd, 3rd favourable, 1st not	\(\frac{2\cdot4\cdot3}{343}=\frac{24}{343}\)
Total	\(\frac{209}{343}\)

Answer: \(\dfrac{209}{343}\).

Q26

Two sets of candidates are competing for the position on the board of directors of a company. The probabilities that the first and second sets will win are 0.6 and 0.4 respectively. If the first set wins, the probability of introducing a new product is 0.8, and the corresponding probability if the second set wins is 0.3. What is the probability that the new product will be introduced?

Solution

Let \(E_1, E_2\) be the events that the first and second sets win, with \(P(E_1)=0.6\) and \(P(E_2)=0.4\). Let \(N\) be the event that the new product is introduced. Then \(P(N/E_1)=0.8\) and \(P(N/E_2)=0.3\). By the theorem of total probability,

\[P(N)=P(E_1)P(N/E_1)+P(E_2)P(N/E_2)=0.6\times0.8+0.4\times0.3=0.48+0.12=0.60.\]

Answer: \(0.6\).

Quick recap: \(P(A/B)=\dfrac{P(A\cap B)}{P(B)}\); for independent events \(P(A\cap B)=P(A)P(B)\); \(P(\text{at least one})=1-P(\text{none})\); and if \(E_1,\dots,E_n\) partition the sample space, \(P(A)=\sum P(E_i)P(A/E_i)\).