Miscellaneous Exercise — Probability

1. 13 marksNCERT Miscellaneous Exercise on Chapter 13

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

   Hint. In each case simplify the intersection A and B first.

   Both parts reduce to identifying $A\cap B$ before dividing.

   (i) If $A\subset B$ then every outcome of $A$ lies in $B$, so $A\cap B=A$. Hence $P(B|A)=\frac{P(A)}{P(A)}=1$.

   (ii) If $A\cap B=\phi$ the events cannot occur together, so $P(A\cap B)=0$ and $P(B|A)=\frac{0}{P(A)}=0$.

   The two answers are the extreme values, which makes sense: containment guarantees $B$, while disjointness rules it out.

   Both results follow directly from the definition, since the intersection collapses to a known set in each case.

   ✦ (i) 1 (ii) 0

2. 23 marksNCERT Miscellaneous Exercise on Chapter 13

   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.

   Hint. Write the sample space as ordered pairs; the two conditions restrict it to different numbers of outcomes.

   Taking the outcomes as ordered pairs (elder, younger) gives $\{BB,BG,GB,GG\}$, all equally likely.

   (i) 'At least one male' leaves $\{BB,BG,GB\}$, which is $3$ outcomes, of which only $BB$ has both male. So the probability is $\frac{1}{3}$.

   (ii) 'The elder child is a female' fixes the first position, leaving $\{GB,GG\}$, which is $2$ outcomes, of which only $GG$ has both female. So the probability is $\frac{1}{2}$.

   Specifying which child is female is stronger information than knowing merely that one of them is, which is why the second answer is larger.

   ✦ (i) 1/3 (ii) 1/2

3. 34 marksNCERT Miscellaneous Exercise on Chapter 13

   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 numbers of males and females.

   Hint. Equal numbers means both priors are 1/2, so they cancel from the Bayes ratio.

   Equal numbers of men and women make both priors $\frac{1}{2}$, so they cancel entirely.

   Let $M$ be 'male' and $G$ be 'has grey hair', with $P(G|M)=0.05$ and $P(G|M^{\prime })=0.0025$.

   $P(M|G)=\frac{0.5\times 0.05}{0.5\times 0.05+0.5\times 0.0025}=\frac{0.05}{0.05+0.0025}=\frac{0.05}{0.0525}=\frac{500}{525}=\frac{20}{21}$

   The answer is close to $1$ because grey hair is twenty times more common in men under these assumptions.

   ✦ P(male | grey haired) = 20/21

4. 45 marksNCERT Miscellaneous Exercise on Chapter 13

   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?

   Hint. This needs the binomial distribution, with n = 10 and p = 9/10. 'At most 6' means summing the terms for 0 through 6.

   This question requires the binomial distribution, a topic the current edition of the chapter no longer teaches anywhere — see the chapter page for the full note on this.

   Treat each person as an independent trial with $n=10$ and $p=\frac{9}{10}$ for right-handedness.

   'At most 6 are right-handed' means $X\le 6$, so the required probability is:

   $P(X\le 6)={\sum }_{r=0}^6\left(\genfrac{}{}{0ex}{}{10}{r}\right){\left(\frac{9}{10}\right)}^r{\left(\frac{1}{10}\right)}^{10-r}$

   Evaluating the seven terms and adding gives $\frac{7996999}{625000000}\approx 0.0128$.

   The value is small because $p$ is high, so having $6$ or fewer right-handers out of $10$ is a considerable departure from the expected $9$.

   ✦ P(X <= 6) = sum from r=0 to 6 of C(10,r)(9/10)^r(1/10)^(10-r), approximately 0.0128

5. 54 marksNCERT Miscellaneous Exercise on Chapter 13

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

   Hint. A leap year is 366 days. Work out how many complete weeks that is and what is left over.

   The question turns on the remainder after removing whole weeks.

   A leap year has $366$ days, and $366=52\times 7+2$, so it contains $52$ complete weeks plus $2$ extra days.

   Every day of the week therefore occurs at least $52$ times, and a 53rd Tuesday requires one of the two extra days to be a Tuesday.

   The two extra days are consecutive, giving $7$ equally likely possibilities: (Sun,Mon), (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun).

   Of these, (Mon,Tue) and (Tue,Wed) contain a Tuesday, which is $2$ cases.

   Therefore the probability is $\frac{2}{7}$.

   ✦ P = 2/7

6. 65 marksNCERT Miscellaneous Exercise on Chapter 13

   Suppose we have four boxes A, B, C and D containing coloured marbles: A has 1 red, 6 white, 3 black; B has 6 red, 2 white, 2 black; C has 8 red, 1 white, 1 black; D has 0 red, 6 white, 4 black. One of the boxes is 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?

   Hint. Every box holds 10 marbles, which keeps the conditional probabilities simple. Box D can contribute nothing.

   Each box contains exactly $10$ marbles, so the conditional probabilities are read straight off the red counts.

   With each box equally likely at $\frac{1}{4}$, and $R$ denoting a red marble:

   $P(R|A)=\frac{1}{10}$, $P(R|B)=\frac{6}{10}$, $P(R|C)=\frac{8}{10}$, $P(R|D)=0$.

   The denominator is $\frac{1}{4}\left(\frac{1+6+8+0}{10}\right)=\frac{1}{4}\times \frac{15}{10}=\frac{3}{8}$.

   The common factor $\frac{1}{4}$ cancels, so each answer is simply that box's red count divided by $15$:

   $P(A|R)=\frac{1}{15}$, $P(B|R)=\frac{6}{15}=\frac{2}{5}$, $P(C|R)=\frac{8}{15}$.

   As a check these sum to $1$, which they must, since box D can never yield a red marble.

   ✦ P(A|red) = 1/15, P(B|red) = 2/5, P(C|red) = 8/15

7. 75 marksNCERT Miscellaneous Exercise on Chapter 13

   Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of a 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.

   Hint. 'Reduces the risk by 30%' means the risk becomes 70% of 0.4, not 0.4 minus 0.30.

   The phrase 'reduces the risk by 30%' means the risk is multiplied by $0.7$, not decreased by $0.30$ in absolute terms — this is the step that decides the question.

   Let $E_1$ be 'follows yoga' and $E_2$ 'takes the drug', each with probability $\frac{1}{2}$.

   With $A$ denoting a heart attack, $P(A|E_1)=0.40\times 0.70=0.28$ and $P(A|E_2)=0.40\times 0.75=0.30$.

   $P(E_1|A)=\frac{0.5\times 0.28}{0.5\times 0.28+0.5\times 0.30}=\frac{0.28}{0.58}=\frac{14}{29}$

   The answer is just under $\frac{1}{2}$, reflecting that yoga is the slightly more effective of the two options.

   Yoga wins narrowly because a 30 percent reduction leaves a smaller residual risk than a 25 percent one.

   ✦ P(yoga | heart attack) = 14/29

8. 84 marksNCERT Miscellaneous Exercise on Chapter 13

   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 are chosen independently, each value being assumed with probability $\frac{1}{2}$.

   Hint. Write the determinant as ad - bc and ask what makes it strictly greater than zero when every entry is 0 or 1.

   Expanding the determinant reduces the problem to a small counting exercise.

   For $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|$ the value is $ad-bc$, and each of the four entries is $0$ or $1$, giving $2^4=16$ equally likely determinants.

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

   Now $ad=1$ forces $a=d=1$, which is $1$ way.

   And $bc=0$ means $b$ and $c$ are not both $1$, which is $3$ of the $4$ combinations.

   So there are $1\times 3=3$ favourable cases, giving probability $\frac{3}{16}$.

   ✦ P(determinant is positive) = 3/16

9. 94 marksNCERT Miscellaneous Exercise on Chapter 13

   An electronic assembly consists of two subsystems A and B. From previous testing procedures, the following probabilities are known: $P$(A fails)$=0.2$, $P$(B fails alone)$=0.15$, $P$(A and B fail)$=0.15$. Evaluate (i) $P$(A fails|B has failed) (ii) $P$(A fails alone).

   Hint. 'B fails alone' excludes the case where A also fails, so P(B fails) is the sum of two pieces.

   The phrase 'fails alone' excludes the joint failure, which is the key to assembling $P(B\text{ fails})$.

   Since B failing splits into B alone and both failing:

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

   (i) $P(A\text{ fails}|B\text{ has failed})=\frac{P(A\cap B)}{P(B)}=\frac{0.15}{0.30}=\frac{1}{2}$

   (ii) $P(A\text{ fails alone})=P(A\text{ fails})-P(A\text{ and }B\text{ fail})=0.2-0.15=0.05$

   ✦ (i) 1/2 (ii) 0.05

10. 105 marksNCERT Miscellaneous Exercise on Chapter 13

    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.

    Hint. Split on the colour of the transferred ball; Bag II holds 10 balls after the transfer in either case.

    The colour of the transferred ball changes Bag II's composition, so it forms the partition.

    Let $E_1$ be 'a red ball is transferred', with $P(E_1)=\frac{3}{7}$, and $E_2$ be 'a black ball is transferred', with $P(E_2)=\frac{4}{7}$.

    After the transfer Bag II holds $10$ balls either way.

    Under $E_1$ it has $5$ red and $5$ black, so $P(R|E_1)=\frac{5}{10}=\frac{1}{2}$.

    Under $E_2$ it has $4$ red and $6$ black, so $P(R|E_2)=\frac{4}{10}=\frac{2}{5}$.

    $P(E_2|R)=\frac{\frac{4}{7}\times \frac{2}{5}}{\frac{3}{7}\times \frac{1}{2}+\frac{4}{7}\times \frac{2}{5}}=\frac{\frac{8}{35}}{\frac{3}{14}+\frac{8}{35}}$

    Putting the denominator over $70$ gives $\frac{15}{70}+\frac{16}{70}=\frac{31}{70}$, while the numerator is $\frac{16}{70}$.

    Therefore $P(E_2|R)=\frac{16}{31}$.

    ✦ P(transferred ball is black | red drawn) = 16/31

11. 112 marksNCERT Miscellaneous Exercise on Chapter 13

    Choose the correct answer. If A and B are two events such that $P(A)\ne 0$ and $P(B|A)=1$, then (A) $A\subset B$ (B) $B\subset A$ (C) $B=\phi$ (D) $A=\phi$

    Hint. Translate P(B|A) = 1 into a statement about P(A intersect B).

    Converting the conditional statement into an intersection makes the containment visible.

    $P(B|A)=1$ means $\frac{P(A\cap B)}{P(A)}=1$, so $P(A\cap B)=P(A)$.

    This says that all of $A$'s probability lies inside $B$, so no part of $A$ falls outside $B$.

    Hence $A\subset B$, which is option (A).

    In words: given that $A$ has occurred, $B$ is certain, which can only happen if $A$ is contained in $B$.

    ✦ (A) A is a subset of B

12. 122 marksNCERT Miscellaneous Exercise on Chapter 13

    Choose the correct answer. If $P(A|B)>P(A)$, then which of the following is correct? (A) $P(B|A)<P(B)$ (B) $P(A\cap B)<P(A)\cdot P(B)$ (C) $P(B|A)>P(B)$ (D) $P(B|A)=P(B)$

    Hint. Turn the given inequality into a statement about the intersection, then divide by P(A) instead.

    Rewriting the inequality in terms of the intersection lets it be re-divided the other way.

    From $P(A|B)>P(A)$ we get $\frac{P(A\cap B)}{P(B)}>P(A)$, hence $P(A\cap B)>P(A)\cdot P(B)$.

    This immediately rules out option (B), which claims the opposite inequality.

    Now divide both sides by $P(A)$ instead:

    $\frac{P(A\cap B)}{P(A)}>P(B)$, that is $P(B|A)>P(B)$.

    So option (C) is correct. The relation is symmetric: if $B$ makes $A$ more likely, then $A$ makes $B$ more likely.

    ✦ (C) P(B|A) > P(B)

13. 132 marksNCERT Miscellaneous Exercise on Chapter 13

    Choose the correct answer. If A and B are any two events such that $P(A)+P(B)-P(A\text{ and }B)=P(A)$, then (A) $P(B|A)=1$ (B) $P(A|B)=1$ (C) $P(B|A)=0$ (D) $P(A|B)=0$

    Hint. Cancel P(A) from both sides and see what remains.

    Cancelling the common term reduces the equation to a single relation.

    The left-hand side is $P(A\cup B)$ by the addition rule, so the condition says $P(A\cup B)=P(A)$.

    Cancelling $P(A)$ from both sides of $P(A)+P(B)-P(A\cap B)=P(A)$ leaves $P(B)-P(A\cap B)=0$, that is $P(A\cap B)=P(B)$.

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

    So option (B) is correct; equivalently $B\subset A$.

    ✦ (B) P(A|B) = 1

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.