NCERT Class 12 Mathematics — Probability, Miscellaneous Exercise on Chapter 13. All 13 questions solved.

Thirteen questions drawing on the whole chapter: conditional probability, independence, the theorem of total probability, and Bayes’ theorem. Nothing tells you which is which.

Key insight. Two structural questions decide almost every problem here.

Is anything being inferred backwards? If the question gives you an outcome and asks which cause produced it — questions 3, 6, 7 and 10 — it is Bayes’ theorem. If it just asks for the probability of an outcome, it is total probability or a direct count.

Are the trials independent? Questions 4 and 8 are counting problems where each item is chosen independently, so the binomial pattern $\binom{n}{r}p^r(1-p)^{n-r}$ applies. Question 5 is a pure counting question about the calendar with no independence at all.

Question 1

$\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A}) \ne 0$. Find $\mathrm{P}(\mathrm{B}|\mathrm{A})$, if (i) $\mathrm{A}$ is a subset of $\mathrm{B}$ (ii) $\mathrm{A} \cap \mathrm{B} = \phi$.

Solution.

(i) If $\mathrm{A} \subset \mathrm{B}$ then every outcome of $\mathrm{A}$ is also in $\mathrm{B}$, so $\mathrm{A} \cap \mathrm{B} = \mathrm{A}$ and

$$\mathrm{P}(\mathrm{B}|\mathrm{A}) = \frac{\mathrm{P}(\mathrm{A})}{\mathrm{P}(\mathrm{A})} = 1$$

(ii) If the events are disjoint then $\mathrm{P}(\mathrm{A} \cap \mathrm{B}) = 0$, so the conditional probability is $0$.

The two parts are the extremes: given $\mathrm{A}$, the event $\mathrm{B}$ is either certain or impossible.

Question 2

A couple has two children,

(i) Find the probability that both children are males, if it is known that at least one of the children is male.

(ii) Find the probability that both children are females, if it is known that the elder child is a female.

Solution. Sample space $\{\mathrm{MM, MF, FM, FF}\}$, elder child first.

(i) “At least one male” is $\{\mathrm{MM, MF, FM}\}$ — three outcomes, one of which is $\mathrm{MM}$.

(ii) “Elder is female” is $\{\mathrm{FM, FF}\}$ — two outcomes, one of which is $\mathrm{FF}$.

Naming which child is known changes the answer from $\tfrac13$ to $\tfrac12$.

Question 3

Suppose that $5\%$ of men and $0.25\%$ of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Solution. Causes: male ($\tfrac12$), female ($\tfrac12$). Evidence: grey hair. The equal priors cancel:

$$\mathrm{P}(\text{male}|\text{grey}) = \frac{0.05}{0.05 + 0.0025} = \frac{0.05}{0.0525} = \frac{500}{525}$$

Question 4

Suppose that $90\%$ of people are right-handed. What is the probability that at most $6$ of a random sample of $10$ people are right-handed?

Solution. Each person is independently right-handed with probability $p = 0.9$, so the number of right-handers among $10$ follows the binomial pattern.

“At most $6$” would need seven terms, so use the complement — “at least $7$” needs only four:

$$\mathrm{P}(\text{at most } 6) = 1 – \sum_{r=7}^{10}\binom{10}{r}(0.9)^r(0.1)^{10-r}$$

NCERT leaves the answer in this form; evaluated, it is about $0.0128$.

Question 5

If a leap year is selected at random, what is the chance that it will contain $53$ Tuesdays?

Solution. A leap year has $366$ days, which is $52$ complete weeks plus $2$ extra days. Every day of the week therefore occurs at least $52$ times, and the two extra consecutive days decide which get a $53$rd.

The two extra days form one of seven equally likely consecutive pairs:

$$\{\text{Sun, Mon}\},\ \{\text{Mon, Tue}\},\ \{\text{Tue, Wed}\},\ \{\text{Wed, Thu}\},\ \{\text{Thu, Fri}\},\ \{\text{Fri, Sat}\},\ \{\text{Sat, Sun}\}$$

Tuesday appears in exactly two of them.

Question 6

Suppose we have four boxes A, B, C and D containing coloured marbles as given below. One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A? box B? box C?

Box	Red	White	Black
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

Solution. Every box holds $10$ marbles, so the likelihoods are simply the red counts over $10$, and each box has prior $\tfrac14$. The common factor $\tfrac{1}{40}$ cancels throughout, leaving the red counts themselves:

$$\text{total} = 1 + 6 + 8 + 0 = 15$$

$$\mathrm{P}(\mathrm{A}|\text{red}) = \frac{1}{15}, \qquad \mathrm{P}(\mathrm{B}|\text{red}) = \frac{6}{15} = \frac25, \qquad \mathrm{P}(\mathrm{C}|\text{red}) = \frac{8}{15}$$

Box D contributes nothing because it has no red marbles — but it still had to be included as a possible cause, and the fact that its likelihood is zero is what removes it.

Question 7

Assume that the chances of a patient having a heart attack is $40\%$. It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30\%$ and prescription of certain drug reduces its chances by $25\%$. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga.

Solution. Causes: yoga ($\tfrac12$) or drug ($\tfrac12$). Evidence: a heart attack occurred.

“Reduces the risk by $30\%$” means the $0.4$ chance becomes $0.4 \times 0.7 = 0.28$; “by $25\%$” makes it $0.4 \times 0.75 = 0.30$.

$$\mathrm{P}(\text{yoga}|\text{attack}) = \frac{\frac12 \times 0.28}{\frac12 \times 0.28 + \frac12 \times 0.30} = \frac{0.28}{0.58}$$

The phrase “reduces by $30\%$” means a relative reduction, not a subtraction of $0.3$ — a $0.4 – 0.3 = 0.1$ reading gives the wrong answer.

Question 8

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\tfrac12$.)

Solution. The determinant is $\begin{vmatrix}a & b \\ c & d\end{vmatrix} = ad – bc$, with each of $a, b, c, d$ independently $0$ or $1$ — so $2^4 = 16$ equally likely determinants.

Since $ad$ and $bc$ are each $0$ or $1$, the value $ad – bc$ is positive only when $ad = 1$ and $bc = 0$.

• $ad = 1$ forces $a = d = 1$: exactly one way.
• $bc = 0$ excludes only $b = c = 1$: three ways out of four.

So there are $1 \times 3 = 3$ favourable determinants.

Question 9

An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

$$\mathrm{P}(\mathrm{A}\text{ fails}) = 0.2, \qquad \mathrm{P}(\mathrm{B}\text{ fails alone}) = 0.15, \qquad \mathrm{P}(\mathrm{A}\text{ and }\mathrm{B}\text{ fail}) = 0.15$$

Evaluate (i) $\mathrm{P}(\mathrm{A}$ fails$|\mathrm{B}$ has failed$)$ (ii) $\mathrm{P}(\mathrm{A}$ fails alone$)$.

Solution. The phrase “fails alone” means that subsystem fails and the other does not. So

$$\mathrm{P}(\mathrm{B}\text{ fails}) = \mathrm{P}(\mathrm{B}\text{ fails alone}) + \mathrm{P}(\text{both fail}) = 0.15 + 0.15 = 0.30$$

(i)

$$\mathrm{P}(\mathrm{A}\text{ fails}|\mathrm{B}\text{ has failed}) = \frac{0.15}{0.30} = 0.5$$

(ii)

$$\mathrm{P}(\mathrm{A}\text{ fails alone}) = \mathrm{P}(\mathrm{A}\text{ fails}) – \mathrm{P}(\text{both fail}) = 0.2 – 0.15 = 0.05$$

Question 10

Bag I contains $3$ red and $4$ black balls and Bag II contains $4$ red and $5$ black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Solution. Causes: the transferred ball was red ($\tfrac37$) or black ($\tfrac47$). Evidence: a red ball is drawn from Bag II, which now holds $10$ balls.

If a red was transferred, Bag II has $5$ red out of $10$; if a black, $4$ red out of $10$.

$$\mathrm{P}(\text{black transferred}|\text{red drawn}) = \frac{\frac47 \cdot \frac{4}{10}}{\frac37 \cdot \frac{5}{10} + \frac47 \cdot \frac{4}{10}} = \frac{16}{15 + 16}$$

Choose the correct answer in each of the following.

Question 11

If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A}) \ne 0$ and $\mathrm{P}(\mathrm{B}|\mathrm{A}) = 1$, then

Solution. $\mathrm{P}(\mathrm{B}|\mathrm{A}) = 1$ means

$$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} = 1 \quad\Longrightarrow\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \mathrm{P}(\mathrm{A})$$

So $\mathrm{A}$ contributes nothing outside $\mathrm{B}$: whenever $\mathrm{A}$ occurs, so does $\mathrm{B}$. That is exactly $\mathrm{A} \subset \mathrm{B}$.

Question 12

If $\mathrm{P}(\mathrm{A}|\mathrm{B}) > \mathrm{P}(\mathrm{A})$, then which of the following is correct:

Solution. The hypothesis says

$$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} > \mathrm{P}(\mathrm{A}) \quad\Longrightarrow\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B}) > \mathrm{P}(\mathrm{A})\,\mathrm{P}(\mathrm{B})$$

which immediately rules out (B). Dividing that same inequality by $\mathrm{P}(\mathrm{A})$ instead gives

$$\frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{A})} > \mathrm{P}(\mathrm{B}) \quad\Longrightarrow\quad \mathrm{P}(\mathrm{B}|\mathrm{A}) > \mathrm{P}(\mathrm{B})$$

The relation is symmetric: if $\mathrm{B}$ makes $\mathrm{A}$ more likely, then $\mathrm{A}$ makes $\mathrm{B}$ more likely.

Question 13

If $\mathrm{A}$ and $\mathrm{B}$ are any two events such that $\mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B}) – \mathrm{P}(\mathrm{A}$ and $\mathrm{B}) = \mathrm{P}(\mathrm{A})$, then

Solution. Cancelling $\mathrm{P}(\mathrm{A})$ from both sides:

$$\mathrm{P}(\mathrm{B}) – \mathrm{P}(\mathrm{A} \cap \mathrm{B}) = 0 \quad\Longrightarrow\quad \mathrm{P}(\mathrm{A} \cap \mathrm{B}) = \mathrm{P}(\mathrm{B})$$

Therefore

$$\mathrm{P}(\mathrm{A}|\mathrm{B}) = \frac{\mathrm{P}(\mathrm{A} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})} = 1$$

Note this is the mirror image of question 11: there $\mathrm{A} \subset \mathrm{B}$, here $\mathrm{B} \subset \mathrm{A}$. Which conditional equals $1$ depends on which event is contained in which.

Common mistakes

• Reading “reduces the risk by $30\%$” as subtracting $0.3$. Question 7 needs $0.4 \times 0.7$, not $0.4 – 0.3$.
• Confusing “fails” with “fails alone”. Question 9 supplies $\mathrm{P}(\mathrm{B}$ fails alone$)$ and asks for $\mathrm{P}(\mathrm{A}$ fails alone$)$; the intersection must be added or subtracted correctly each time.
• Dropping a cause with zero likelihood. Box D in question 6 must appear in the setup even though it contributes nothing; forgetting it is harmless here but a habit worth avoiding.
• Working “at most $6$” directly. Question 4’s complement has four terms instead of seven.
• Assuming a leap year’s extra days are independent. In question 5 they are two consecutive days, which is why there are seven pairs, not $49$.
• Getting the containment backwards. Questions 11 and 13 are mirror images; read carefully which conditional probability equals $1$.

Practise next

• Exercise 13.3 — Bayes’ theorem drilled one question at a time.
• Exercise 13.1 — the conditional probability definitions that questions 1, 11, 12 and 13 rest on.