NCERT Class 12 Mathematics — Probability, Exercise 13.3. All 14 questions solved.

Every question but the last two is Bayes’ theorem:

$$\mathrm{P}(\mathrm{E}_i\,|\,\mathrm{A}) = \frac{\mathrm{P}(\mathrm{E}_i)\,\mathrm{P}(\mathrm{A}|\mathrm{E}_i)}{\displaystyle\sum_j \mathrm{P}(\mathrm{E}_j)\,\mathrm{P}(\mathrm{A}|\mathrm{E}_j)}$$

where $\mathrm{E}_1, \mathrm{E}_2, \ldots$ partition the sample space and $\mathrm{A}$ is the observed event. The denominator is the theorem of total probability — the probability of $\mathrm{A}$ however it came about.

Key insight. Set every one of these questions out the same way before computing anything:

1. Name the causes $\mathrm{E}_1, \mathrm{E}_2, \ldots$ — the hypotheses you want to choose between. These are what the question asks about.
2. Name the evidence $\mathrm{A}$ — the thing that has already been observed. In almost every question here, this is the second-mentioned fact.
3. Write the priors $\mathrm{P}(\mathrm{E}_i)$ and the likelihoods $\mathrm{P}(\mathrm{A}|\mathrm{E}_i)$.

The single commonest error in the exercise is swapping steps 1 and 2. In question 3, “the student has an A grade” is the evidence and “the student is a hostler” is the cause — not the other way round.

Question 1

An urn contains $5$ red and $5$ black balls. A ball is drawn at random, its colour is noted and 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?

Solution. This one uses the theorem of total probability, not Bayes — nothing is being inferred backwards.

After the first draw the urn holds $12$ balls. If the first was red there are now $7$ red; if black, $5$ red.

$$\mathrm{P}(\text{2nd red}) = \frac12 \cdot \frac{7}{12} + \frac12 \cdot \frac{5}{12} = \frac{7 + 5}{24}$$

The answer is the same as the original proportion of red balls, which is not a coincidence — adding balls of whichever colour was drawn is a symmetric process.

Question 2

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.

Solution. Causes: $\mathrm{E}_1$ (bag I chosen), $\mathrm{E}_2$ (bag II chosen), each with prior $\tfrac12$. Evidence $\mathrm{A}$: the ball is red.

$$\mathrm{P}(\mathrm{A}|\mathrm{E}_1) = \frac48 = \frac12, \qquad \mathrm{P}(\mathrm{A}|\mathrm{E}_2) = \frac28 = \frac14$$

$$\mathrm{P}(\mathrm{E}_1|\mathrm{A}) = \frac{\frac12 \cdot \frac12}{\frac12 \cdot \frac12 + \frac12 \cdot \frac14} = \frac{\frac14}{\frac38}$$

Question 3

Of the students in a college, it is known that $60\%$ 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 in their annual examination. 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?

Solution. Causes: hostler ($0.6$) and day scholar ($0.4$). Evidence: A grade.

$$\mathrm{P}(\text{A}|\text{hostler}) = 0.3, \qquad \mathrm{P}(\text{A}|\text{day scholar}) = 0.2$$

$$\mathrm{P}(\text{hostler}|\text{A}) = \frac{0.6 \times 0.3}{0.6 \times 0.3 + 0.4 \times 0.2} = \frac{0.18}{0.26}$$

Question 4

In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\dfrac34$ be the probability that he knows the answer and $\dfrac14$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\dfrac14$, what is the probability that the student knows the answer given that he answered it correctly?

Solution. Causes: knows ($\tfrac34$), guesses ($\tfrac14$). Evidence: correct.

A student who knows is correct with probability $1$; one who guesses, with probability $\tfrac14$.

$$\mathrm{P}(\text{knows}|\text{correct}) = \frac{\frac34 \times 1}{\frac34 \times 1 + \frac14 \times \frac14} = \frac{\frac34}{\frac{13}{16}}$$

Question 5

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 person 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?

Solution. Causes: has the disease ($0.001$), healthy ($0.999$). Evidence: positive test.

$$\mathrm{P}(+|\text{disease}) = 0.99, \qquad \mathrm{P}(+|\text{healthy}) = 0.005$$

$$\mathrm{P}(\text{disease}|+) = \frac{0.001 \times 0.99}{0.001 \times 0.99 + 0.999 \times 0.005} = \frac{0.00099}{0.005985}$$

The answer is under $17\%$, even though the test is $99\%$ effective — because the disease is so rare that the false positives from the healthy majority outnumber the true positives. This is the classic lesson of Bayes’ theorem.

Question 6

There are three coins. One is a two headed coin, another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Solution. Causes: the three coins, each with prior $\tfrac13$. Evidence: heads.

$$\mathrm{P}(\mathrm{H}|\text{two-headed}) = 1, \quad \mathrm{P}(\mathrm{H}|\text{biased}) = \frac34, \quad \mathrm{P}(\mathrm{H}|\text{fair}) = \frac12$$

The common factor $\tfrac13$ cancels from every term:

$$\mathrm{P}(\text{two-headed}|\mathrm{H}) = \frac{1}{1 + \frac34 + \frac12} = \frac{1}{\frac94}$$

Question 7

An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability 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?

Solution. The total insured is $12000$, so the priors are $\tfrac{2000}{12000} = \tfrac16$, $\tfrac13$ and $\tfrac12$. It is easier to keep the raw counts, since the $12000$ cancels:

$$\mathrm{P}(\text{scooter}|\text{accident}) = \frac{2000 \times 0.01}{2000 \times 0.01 + 4000 \times 0.03 + 6000 \times 0.15} = \frac{20}{20 + 120 + 900}$$

Question 8

A factory has two machines A and B. Past record shows that machine A produced $60\%$ of the items of output and machine B produced $40\%$ of the items. Further, $2\%$ of the items produced by machine A and $1\%$ produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Solution. Causes: machine A ($0.6$), machine B ($0.4$). Evidence: defective.

$$\mathrm{P}(\mathrm{B}|\text{defective}) = \frac{0.4 \times 0.01}{0.6 \times 0.02 + 0.4 \times 0.01} = \frac{0.004}{0.016}$$

Note that machine B is asked about, so its term goes on top; the denominator always contains every cause.

Question 9

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the 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.

Solution. Causes: first group wins ($0.6$), second group wins ($0.4$). Evidence: a new product was introduced.

$$\mathrm{P}(\text{2nd}|\text{new product}) = \frac{0.4 \times 0.3}{0.6 \times 0.7 + 0.4 \times 0.3} = \frac{0.12}{0.54}$$

Question 10

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?

Solution. Causes: die shows $5$ or $6$ (prior $\tfrac13$), or $1$ to $4$ (prior $\tfrac23$). Evidence: exactly one head.

With three tosses, exactly one head has probability $\tfrac{3}{8}$. With one toss, exactly one head means a head: probability $\tfrac12$.

$$\mathrm{P}(1\text{–}4 \mid \text{one head}) = \frac{\frac23 \cdot \frac12}{\frac13 \cdot \frac38 + \frac23 \cdot \frac12} = \frac{\frac13}{\frac{11}{24}}$$

Question 11

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 is on the job for $30\%$ of the time and C is on the job for $20\%$ of the time. A defective item is produced, what is the probability that it was produced by A?

Solution. Priors $0.5$, $0.3$, $0.2$; likelihoods $0.01$, $0.05$, $0.07$.

$$\mathrm{P}(\mathrm{A}|\text{defective}) = \frac{0.5 \times 0.01}{0.5 \times 0.01 + 0.3 \times 0.05 + 0.2 \times 0.07} = \frac{0.005}{0.005 + 0.015 + 0.014}$$

Question 12

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.

Solution. Causes: the lost card is a diamond (prior $\tfrac{13}{52} = \tfrac14$) or is not (prior $\tfrac34$). Evidence: two diamonds drawn from the remaining $51$.

If the lost card was a diamond, $12$ diamonds remain among $51$ cards:

$$\mathrm{P}(\mathrm{A}|\text{diamond lost}) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{66}{1275}$$

If not, $13$ diamonds remain:

$$\mathrm{P}(\mathrm{A}|\text{non-diamond lost}) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{78}{1275}$$

The common denominator $\binom{51}{2}$ cancels:

$$\mathrm{P}(\text{diamond lost}|\mathrm{A}) = \frac{\frac14 \times 66}{\frac14 \times 66 + \frac34 \times 78} = \frac{16.5}{75}$$

Question 13

Probability that A speaks truth is $\dfrac45$. A coin is tossed. A reports that a head appears. The probability that actually there was head is

Solution. Causes: the coin actually showed a head ($\tfrac12$) or a tail ($\tfrac12$). Evidence: A reports a head.

A reports a head when the coin was a head and he tells the truth ($\tfrac45$), or when it was a tail and he lies ($\tfrac15$).

$$\mathrm{P}(\text{head}|\text{reports head}) = \frac{\frac12 \cdot \frac45}{\frac12 \cdot \frac45 + \frac12 \cdot \frac15} = \frac{\frac45}{1}$$

The equal priors make the answer equal to A’s truthfulness — but only because the coin is fair.

Question 14

If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{A} \subset \mathrm{B}$ and $\mathrm{P}(\mathrm{B}) \ne 0$, then which of the following is correct?

Solution. Since $\mathrm{A} \subset \mathrm{B}$, we have $\mathrm{A} \cap \mathrm{B} = \mathrm{A}$, so

$$\mathrm{P}(\mathrm{A}|\mathrm{B}) = \frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{B})}$$

And $\mathrm{P}(\mathrm{B}) \le 1$, so dividing by it can only increase the value (or leave it unchanged when $\mathrm{P}(\mathrm{B}) = 1$). Hence $\mathrm{P}(\mathrm{A}|\mathrm{B}) \ge \mathrm{P}(\mathrm{A})$.

Option (A) has the fraction upside down; option (B) has the inequality backwards.

Common mistakes

• Swapping cause and evidence. The event you are asked about goes on the left of the bar in the answer; the event that has happened goes on the right.
• Leaving a cause out of the denominator. The denominator is the total probability of the evidence, summed over every hypothesis — question 11 has three terms, question 7 has three, question 6 has three.
• Assuming the answer must be large because the likelihood is. Question 5 has a $99\%$ effective test and a posterior of about $16\%$, because the prior is tiny.
• Using the wrong denominator after a card is removed. In question 12 the draws are from $51$ cards, not $52$.
• Forgetting that the priors need not be equal. They are equal in questions 2 and 6, and unequal in almost every other question here.
• Computing “exactly one head” as $\tfrac12$ for three tosses. In question 10 it is $\tfrac38$; only for a single toss is it $\tfrac12$.

Practise next

• Miscellaneous Exercise on Chapter 13 — Bayes’ theorem alongside the earlier material, with no signposting.
• Exercise 13.2 — the multiplication theorem that supplies every likelihood used here.