Exercise 13.3 — Probability

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14 marksNCERT Exercise 13.3
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and it is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Hint. Split on the colour of the first ball, since it determines the composition of the urn for the second draw.
The composition of the urn for the second draw depends on the first outcome, so the theorem of total probability applies.

If the first ball is red (probability $\frac{1}{2}$ ), the urn becomes $7$ red and $5$ black, a total of $12$ , so the second is red with probability $\frac{7}{12}$ .

If the first ball is black (probability $\frac{1}{2}$ ), the urn becomes $5$ red and $7$ black, so the second is red with probability $\frac{5}{12}$ .

Adding the weighted contributions:

$P = \frac{1}{2} \times \frac{7}{12} + \frac{1}{2} \times \frac{5}{12} = \frac{7 + 5}{24} = \frac{12}{24} = \frac{1}{2}$

The answer equals the original proportion of red balls, because the procedure treats both colours symmetrically.

✦ P(second ball is red) = 1/2
24 marksNCERT Exercise 13.3
A bag contains 4 red and 4 black balls; another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag, which is found to be red. Find the probability that the ball is drawn from the first bag.
Hint. This asks for a cause given an effect, so use Bayes' theorem with the two bags as the partition.
The question runs backwards from the observed colour to the bag, which is exactly what Bayes' theorem does.

Let $E_{1}, E_{2}$ be the events of choosing bag I and bag II, each with probability $\frac{1}{2}$ , and let $R$ be drawing a red ball.

$P (R ∣ E_{1}) = \frac{4}{8} = \frac{1}{2}$ and $P (R ∣ E_{2}) = \frac{2}{8} = \frac{1}{4}$ .

By Bayes' theorem:

$P (E_{1} ∣ R) = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{8}} = \frac{\frac{1}{4}}{\frac{3}{8}} = \frac{2}{3}$

The answer exceeds $\frac{1}{2}$ because the first bag is richer in red balls, so a red draw is evidence favouring it.

✦ P(first bag | red) = 2/3
34 marksNCERT Exercise 13.3
Of the students in a college, 60% reside in hostel and 40% are day scholars. Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade. At the end of the year, one student is chosen at random from the college and he has an A grade. What is the probability that the student is a hostler?
Hint. Take residence as the partition and 'A grade' as the observed evidence.
Residence forms the partition and the grade is the evidence, so Bayes' theorem gives the reversed probability.

Let $E_{1}$ be 'resides in hostel' with $P (E_{1}) = 0.6$ , and $E_{2}$ be 'day scholar' with $P (E_{2}) = 0.4$ .

Let $A$ be 'attains A grade', with $P (A ∣ E_{1}) = 0.3$ and $P (A ∣ E_{2}) = 0.2$ .

$P (E_{1} ∣ A) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \frac{0.18}{0.18 + 0.08} = \frac{0.18}{0.26} = \frac{9}{13}$

Both the larger population share and the higher A-grade rate push the answer above $0.6$ .

✦ P(hostler | A grade) = 9/13
44 marksNCERT Exercise 13.3
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ the probability that he guesses. Assuming that a student who guesses will be correct with probability $\frac{1}{4}$ , what is the probability that the student knows the answer given that he answered it correctly?
Hint. A student who knows the answer is correct with probability 1 — state that explicitly before applying Bayes.
The step most often left implicit is that knowing the answer guarantees correctness.

Let $E_{1}$ be 'knows' with $P (E_{1}) = \frac{3}{4}$ , and $E_{2}$ be 'guesses' with $P (E_{2}) = \frac{1}{4}$ .

Let $C$ be 'answers correctly'. Then $P (C ∣ E_{1}) = 1$ and $P (C ∣ E_{2}) = \frac{1}{4}$ .

$P (E_{1} ∣ C) = \frac{\frac{3}{4} \times 1}{\frac{3}{4} \times 1 + \frac{1}{4} \times \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{3}{4} + \frac{1}{16}} = \frac{\frac{12}{16}}{\frac{13}{16}} = \frac{12}{13}$

A correct answer raises the probability from $\frac{3}{4}$ to $\frac{12}{13}$ , since guessing rarely produces a correct response.

✦ P(knows | correct) = 12/13
55 marksNCERT Exercise 13.3
A laboratory blood test is 99% effective in detecting a certain disease when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy persons tested. If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Hint. Convert each percentage to a probability carefully, then apply Bayes with 'has disease' and 'healthy' as the partition.
Careful conversion of the percentages is the whole difficulty in this classic question.

Let $D$ be 'has the disease', with $P (D) = 0.001$ and $P (D^{'}) = 0.999$ .

Let $T$ be 'tests positive', with $P (T ∣ D) = 0.99$ and $P (T ∣ D^{'}) = 0.005$ .

$P (D ∣ T) = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.00099 + 0.004995} = \frac{0.00099}{0.005985}$

Multiplying numerator and denominator by $1 0^{6}$ gives $\frac{990}{5985}$ , and dividing both by $45$ gives $\frac{22}{133}$ .

This is only about $0.165$ , which surprises most students: because the disease is so rare, the false positives from the large healthy population outnumber the true positives.

✦ P(disease | positive) = 22/133, approximately 0.165
64 marksNCERT Exercise 13.3
There are three coins. One is a two-headed coin, another is a biased coin that comes up heads 75% of the time, and the third is an unbiased coin. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?
Hint. The two-headed coin shows heads with probability 1; write down all three conditional probabilities before combining.
Each coin has its own probability of showing heads, and the three are equally likely to be chosen.

Let $E_{1}, E_{2}, E_{3}$ be choosing the two-headed, biased and unbiased coin, each with probability $\frac{1}{3}$ .

Let $H$ be 'shows heads', with $P (H ∣ E_{1}) = 1$ , $P (H ∣ E_{2}) = \frac{3}{4}$ and $P (H ∣ E_{3}) = \frac{1}{2}$ .

$P (E_{1} ∣ H) = \frac{\frac{1}{3} \times 1}{\frac{1}{3} ( 1 + \frac{3}{4} + \frac{1}{2} )} = \frac{1}{\frac{9}{4}} = \frac{4}{9}$

The common factor $\frac{1}{3}$ cancels throughout, which is always the case when the partition is uniform.

The two-headed coin dominates the numerator because it is the only coin certain to show heads.

✦ P(two-headed | heads) = 4/9
74 marksNCERT Exercise 13.3
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Hint. Convert the counts into prior probabilities out of the total of 12000 first.
The vehicle counts must first be turned into prior probabilities.

The total insured is $2000 + 4000 + 6000 = 12000$ , giving priors $\frac{2000}{12000} = \frac{1}{6}$ , $\frac{4000}{12000} = \frac{1}{3}$ and $\frac{6000}{12000} = \frac{1}{2}$ .

With $A$ denoting an accident:

$P (scooter ∣ A) = \frac{\frac{1}{6} \times 0.01}{\frac{1}{6} \times 0.01 + \frac{1}{3} \times 0.03 + \frac{1}{2} \times 0.15}$

Multiplying every term by $6$ simplifies this to $\frac{0.01}{0.01 + 0.06 + 0.45} = \frac{0.01}{0.52} = \frac{1}{52}$ .

The answer is small because scooter drivers are both the fewest and the least accident-prone.

✦ P(scooter driver | accident) = 1/52
84 marksNCERT Exercise 13.3
A factory has two machines A and B. Past records show that machine A produced 60% of the items and machine B produced 40%. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All items are put into one stockpile and one item is chosen at random and found to be defective. What is the probability that it was produced by machine B?
Hint. Note the question asks about machine B, so B's term goes in the numerator.
Reading which machine the question asks about is essential, since it decides the numerator.

Let $E_{1}, E_{2}$ be production by machines A and B, with $P (E_{1}) = 0.6$ and $P (E_{2}) = 0.4$ .

Let $D$ be 'defective', with $P (D ∣ E_{1}) = 0.02$ and $P (D ∣ E_{2}) = 0.01$ .

$P (E_{2} ∣ D) = \frac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} = \frac{0.004}{0.012 + 0.004} = \frac{0.004}{0.016} = \frac{1}{4}$

Machine B accounts for only a quarter of the defective items, because it both produces less and produces more reliably.

✦ P(machine B | defective) = 1/4
94 marksNCERT Exercise 13.3
Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7, and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Hint. The question asks about the second group, so its term belongs in the numerator.
Winning the position is the partition, and introducing the new product is the evidence.

Let $E_{1}, E_{2}$ be wins by the first and second group, with $P (E_{1}) = 0.6$ and $P (E_{2}) = 0.4$ .

Let $N$ be 'a new product is introduced', with $P (N ∣ E_{1}) = 0.7$ and $P (N ∣ E_{2}) = 0.3$ .

$P (E_{2} ∣ N) = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.42 + 0.12} = \frac{0.12}{0.54} = \frac{2}{9}$

The second group is less likely to win and less likely to innovate, so both factors reduce its share.

✦ P(second group | new product) = 2/9
105 marksNCERT Exercise 13.3
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?
Hint. Compute the probability of exactly one head separately under each branch; the three-toss branch needs a count of favourable sequences.
The die throw decides which experiment is performed, so it forms the partition.

Let $E_{1}$ be 'threw 5 or 6' with $P (E_{1}) = \frac{2}{6} = \frac{1}{3}$ , and $E_{2}$ be 'threw 1, 2, 3 or 4' with $P (E_{2}) = \frac{4}{6} = \frac{2}{3}$ .

Let $S$ be 'exactly one head'. Under $E_{1}$ the coin is tossed three times, and exactly one head occurs in $3$ of the $8$ sequences, so $P (S ∣ E_{1}) = \frac{3}{8}$ .

Under $E_{2}$ the coin is tossed once, and exactly one head means a head, so $P (S ∣ E_{2}) = \frac{1}{2}$ .

$P (E_{2} ∣ S) = \frac{\frac{2}{3} \times \frac{1}{2}}{\frac{1}{3} \times \frac{3}{8} + \frac{2}{3} \times \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{8} + \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{11}{24}} = \frac{8}{11}$

✦ P(threw 1,2,3 or 4 | exactly one head) = 8/11
114 marksNCERT Exercise 13.3
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B for 30% of the time and C for 20% of the time. A defective item is produced. What is the probability that it was produced by A?
Hint. The time shares are the prior probabilities; the defect rates are the conditional probabilities.
Time on the job supplies the priors and the defect rates supply the likelihoods.

Let $E_{1}, E_{2}, E_{3}$ correspond to operators A, B and C, with priors $0.5$ , $0.3$ and $0.2$ .

Let $D$ be 'defective', with $P (D ∣ E_{1}) = 0.01$ , $P (D ∣ E_{2}) = 0.05$ and $P (D ∣ E_{3}) = 0.07$ .

$P (E_{1} ∣ D) = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07} = \frac{0.005}{0.005 + 0.015 + 0.014} = \frac{0.005}{0.034} = \frac{5}{34}$

Although A works half the time, A contributes only about 15% of the defects, because A's defect rate is the lowest by a wide margin.

✦ P(produced by A | defective) = 5/34
125 marksNCERT Exercise 13.3
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Hint. Split on whether the lost card was a diamond, then compute the chance of drawing two diamonds from the remaining 51 in each case.
The identity of the lost card changes how many diamonds remain, so it forms the partition.

Let $E_{1}$ be 'lost card is a diamond' with $P (E_{1}) = \frac{13}{52} = \frac{1}{4}$ , and $E_{2}$ its complement with $P (E_{2}) = \frac{3}{4}$ .

Let $D$ be 'two diamonds are drawn from the remaining 51'.

Under $E_{1}$ , $12$ diamonds remain, so $P (D ∣ E_{1}) = \frac{( 2 12 )}{( 2 51 )} = \frac{66}{1275}$ .

Under $E_{2}$ , $13$ diamonds remain, so $P (D ∣ E_{2}) = \frac{( 2 13 )}{( 2 51 )} = \frac{78}{1275}$ .

The common denominator $(2 51)$ cancels, leaving:

$P (E_{1} ∣ D) = \frac{\frac{1}{4} \times 66}{\frac{1}{4} \times 66 + \frac{3}{4} \times 78} = \frac{16.5}{16.5 + 58.5} = \frac{16.5}{75} = \frac{11}{50}$

✦ P(lost card is a diamond) = 11/50
133 marksNCERT Exercise 13.3
Choose the correct answer. Probability that A speaks truth is $\frac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was a head is (A) $\frac{4}{5}$ (B) $\frac{1}{2}$ (C) $\frac{1}{5}$ (D) $\frac{2}{5}$
Hint. Note that A reports 'head' either by telling the truth about a head or by lying about a tail.
There are two ways for A to report a head, and both must be accounted for.

Let $E_{1}$ be 'the coin actually shows head' and $E_{2}$ 'tail', each with probability $\frac{1}{2}$ .

Let $R$ be 'A reports a head'. If the coin is a head, A reports it truthfully with probability $\frac{4}{5}$ . If it is a tail, A must lie to report a head, which happens with probability $\frac{1}{5}$ .

$P (E_{1} ∣ R) = \frac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2} \times \frac{4}{5} + \frac{1}{2} \times \frac{1}{5}} = \frac{\frac{4}{5}}{\frac{4}{5} + \frac{1}{5}} = \frac{4}{5}$

The answer coincides numerically with A's truthfulness because the coin is fair, making the two priors equal.

✦ (A) 4/5
143 marksNCERT Exercise 13.3
Choose the correct answer. If A and B are two events such that $A \subset B$ and $P (B) \neq = 0$ , then which of the following is correct? (A) $P (A ∣ B) = \frac{P ( B )}{P ( A )}$ (B) $P (A ∣ B) < P (A)$ (C) $P (A ∣ B) \geq P (A)$ (D) None of these
Hint. If A is contained in B, then the intersection of A and B is just A.
Containment simplifies the intersection immediately.

Since $A \subset B$ , we have $A \cap B = A$ , so $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{P ( A )}{P ( B )}$ .

Because $P (B) \leq 1$ , dividing $P (A)$ by $P (B)$ cannot decrease it, so $\frac{P ( A )}{P ( B )} \geq P (A)$ .

Therefore $P (A ∣ B) \geq P (A)$ , which is option (C). Equality holds only when $P (B) = 1$ .

✦ (C) P(A|B) >= P(A)

Solutions written by the tuition.in editorial team and checked against lemh207.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.

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