Exercise 13.1 — Probability

1. 12 marksNCERT Exercise 13.1

   Given that E and F are events such that $P(E)=0.6$, $P(F)=0.3$ and $P(E\cap F)=0.2$, find $P(E|F)$ and $P(F|E)$.

   Hint. Apply the definition of conditional probability twice, swapping which event is the condition.

   The definition of conditional probability divides the joint probability by the probability of the conditioning event.

   $P(E|F)=\frac{P(E\cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}$

   $P(F|E)=\frac{P(E\cap F)}{P(E)}=\frac{0.2}{0.6}=\frac{1}{3}$

   The two answers differ because the denominators differ, which is why conditional probability is not symmetric.

   ✦ P(E|F) = 2/3 and P(F|E) = 1/3

2. 22 marksNCERT Exercise 13.1

   Compute $P(A|B)$, if $P(B)=0.5$ and $P(A\cap B)=0.32$.

   Hint. Substitute directly into P(A|B) = P(A intersect B) / P(B).

   Only one substitution is needed here, since both required quantities are given outright.

   $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.32}{0.5}=0.64$

   As a fraction this is $\frac{16}{25}$.

   ✦ P(A|B) = 0.64 = 16/25

3. 33 marksNCERT Exercise 13.1

   If $P(A)=0.8$, $P(B)=0.5$ and $P(B|A)=0.4$, find (i) $P(A\cap B)$ (ii) $P(A|B)$ (iii) $P(A\cup B)$.

   Hint. Rearrange the conditional probability formula to get the intersection first; everything else follows from it.

   Rearranging the definition gives the intersection, and that single value then unlocks both remaining parts.

   (i) Since $P(B|A)=\frac{P(A\cap B)}{P(A)}$, we get $P(A\cap B)=P(B|A)\cdot P(A)=0.4\times 0.8=0.32$.

   (ii) $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.32}{0.5}=0.64$

   (iii) By the addition rule, $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.8+0.5-0.32=0.98$.

   ✦ (i) 0.32 (ii) 0.64 (iii) 0.98

4. 43 marksNCERT Exercise 13.1

   Evaluate $P(A\cup B)$, if $2P(A)=P(B)=\frac{5}{13}$ and $P(A|B)=\frac{2}{5}$.

   Hint. Read off P(B) first, halve it for P(A), then use P(A|B) to get the intersection.

   The chained equation $2P(A)=P(B)=\frac{5}{13}$ supplies two probabilities at once.

   So $P(B)=\frac{5}{13}$, and since $2P(A)=\frac{5}{13}$, it follows that $P(A)=\frac{5}{26}$.

   From $P(A|B)=\frac{P(A\cap B)}{P(B)}$ we get $P(A\cap B)=\frac{2}{5}\times \frac{5}{13}=\frac{2}{13}$.

   Therefore $P(A\cup B)=\frac{5}{26}+\frac{5}{13}-\frac{2}{13}=\frac{5}{26}+\frac{10}{26}-\frac{4}{26}=\frac{11}{26}$.

   ✦ P(A union B) = 11/26

5. 53 marksNCERT Exercise 13.1

   If $P(A)=\frac{6}{11}$, $P(B)=\frac{5}{11}$ and $P(A\cup B)=\frac{7}{11}$, find (i) $P(A\cap B)$ (ii) $P(A|B)$ (iii) $P(B|A)$.

   Hint. Use the addition rule backwards to extract the intersection.

   Running the addition rule in reverse is what gives the intersection here.

   (i) $P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac{6}{11}+\frac{5}{11}-\frac{7}{11}=\frac{4}{11}$

   (ii) $P(A|B)=\frac{4/11}{5/11}=\frac{4}{5}$

   (iii) $P(B|A)=\frac{4/11}{6/11}=\frac{2}{3}$

   The common denominator cancels in both divisions, which keeps the arithmetic short.

   ✦ (i) 4/11 (ii) 4/5 (iii) 2/3

6. 64 marksNCERT Exercise 13.1

   Determine $P(E|F)$. A coin is tossed three times, where (i) E : head on third toss, F : heads on first two tosses (ii) E : at least two heads, F : at most two heads (iii) E : at most two tails, F : at least one tail.

   Hint. List the 8 equally likely outcomes, then count how many of F also lie in E.

   With three tosses the sample space has $2^3=8$ equally likely outcomes, so every conditional probability reduces to a count of favourable cases within $F$.

   (i) $F=\{HHH,HHT\}$ has $2$ outcomes; of these only $HHH$ has a head on the third toss. So $P(E|F)=\frac{1}{2}$.

   (ii) $F$ is 'at most two heads', which excludes only $HHH$, giving $7$ outcomes. Within those, 'at least two heads' leaves $HHT,HTH,THH$, which is $3$. So $P(E|F)=\frac{3}{7}$.

   (iii) $F$ is 'at least one tail', excluding only $HHH$, so $7$ outcomes. Within those, 'at most two tails' additionally excludes $TTT$, leaving $6$. So $P(E|F)=\frac{6}{7}$.

   ✦ (i) 1/2 (ii) 3/7 (iii) 6/7

7. 73 marksNCERT Exercise 13.1

   Determine $P(E|F)$. Two coins are tossed once, where (i) E : tail appears on one coin, F : one coin shows head (ii) E : no tail appears, F : no head appears.

   Hint. The sample space is {HH, HT, TH, TT}. In part (ii) check whether the two events can occur together at all.

   Here the sample space is $\{HH,HT,TH,TT\}$, only four outcomes, so both parts are settled by inspection.

   (i) 'Tail appears on one coin' means exactly one tail, giving $E=\{HT,TH\}$. 'One coin shows head' means exactly one head, giving $F=\{HT,TH\}$. Since $E$ and $F$ are the same set, $E\cap F=F$ and $P(E|F)=\frac{2}{2}=1$.

   (ii) 'No tail appears' gives $E=\{HH\}$, while 'no head appears' gives $F=\{TT\}$. These are disjoint, so $E\cap F=\phi$ and $P(E|F)=\frac{0}{1}=0$.

   ✦ (i) 1 (ii) 0

8. 83 marksNCERT Exercise 13.1

   Determine $P(E|F)$. A die is thrown three times, E : 4 appears on the third toss, F : 6 and 5 appear respectively on the first two tosses.

   Hint. Condition F pins down the first two throws completely, leaving only the third free.

   Because $F$ fixes both of the first two throws, the conditioning collapses the problem to a single free throw.

   $F$ consists of the outcomes $(6,5,x)$ where $x$ runs from $1$ to $6$, so $F$ has $6$ outcomes.

   Of these, only $(6,5,4)$ also has a $4$ on the third toss, so $E\cap F$ has $1$ outcome.

   Therefore $P(E|F)=\frac{1}{6}$, which is simply the unconditional probability of throwing a 4 — the first two throws carry no information about the third.

   ✦ P(E|F) = 1/6

9. 93 marksNCERT Exercise 13.1

   Determine $P(E|F)$. Mother, father and son line up at random for a family picture. E : son on one end, F : father in middle.

   Hint. There are 3! arrangements. Check whether every arrangement satisfying F also satisfies E.

   Three people can line up in $3!=6$ equally likely orders, so the counting is small enough to list.

   The arrangements with the father in the middle are $MFS$ and $SFM$, so $F$ has $2$ outcomes.

   In both of them the son occupies an end position, so $E\cap F$ also has $2$ outcomes.

   Therefore $P(E|F)=\frac{2}{2}=1$. Once the father is in the middle, only the mother and son remain for the two ends, so the son is certain to be on one end.

   ✦ P(E|F) = 1

10. 104 marksNCERT Exercise 13.1

    A black and a red die are rolled. (a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. (b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

    Hint. In (a) the black die is fixed, leaving 6 outcomes. In (b) the red die has 3 possible values, leaving 18 outcomes.

    Each part restricts one die, which shrinks the sample space before any counting of the sum begins.

    (a) Given the black die shows $5$, the red die can be $1$ to $6$, so there are $6$ outcomes. A sum greater than $9$ needs $5+\text{red}>9$, that is red $>4$, giving red $\in \{5,6\}$ — $2$ outcomes. So the probability is $\frac{2}{6}=\frac{1}{3}$.

    (b) Given the red die is less than $4$, red $\in \{1,2,3\}$ and black runs $1$ to $6$, so there are $3\times 6=18$ outcomes. For a sum of $8$ the pairs (black, red) are $(5,3)$ and $(6,2)$ — $2$ outcomes. So the probability is $\frac{2}{18}=\frac{1}{9}$.

    ✦ (a) 1/3 (b) 1/9

11. 114 marksNCERT Exercise 13.1

    A fair die is rolled. Consider events $E=\{1,3,5\}$, $F=\{2,3\}$ and $G=\{2,3,4,5\}$. Find (i) $P(E|F)$ and $P(F|E)$ (ii) $P(E|G)$ and $P(G|E)$ (iii) $P((E\cup F)|G)$ and $P((E\cap F)|G)$.

    Hint. Since all outcomes are equally likely, each conditional probability is just the size of the intersection divided by the size of the conditioning set.

    All six faces are equally likely, so every conditional probability reduces to a ratio of set sizes.

    (i) $E\cap F=\{3\}$, so $P(E|F)=\frac{1}{2}$ and $P(F|E)=\frac{1}{3}$.

    (ii) $E\cap G=\{3,5\}$, so $P(E|G)=\frac{2}{4}=\frac{1}{2}$ and $P(G|E)=\frac{2}{3}$.

    (iii) $E\cup F=\{1,2,3,5\}$, and intersecting with $G$ gives $\{2,3,5\}$, so $P((E\cup F)|G)=\frac{3}{4}$.

    Also $E\cap F=\{3\}$, and intersecting with $G$ leaves $\{3\}$, so $P((E\cap F)|G)=\frac{1}{4}$.

    ✦ (i) 1/2, 1/3 (ii) 1/2, 2/3 (iii) 3/4, 1/4

12. 123 marksNCERT Exercise 13.1

    Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

    Hint. Write the sample space as ordered pairs (elder, younger). The two conditions cut it down by different amounts.

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

    (i) 'The youngest is a girl' selects those with a girl in the second position, namely $\{BG,GG\}$ — $2$ outcomes. Only $GG$ has both girls, so the probability is $\frac{1}{2}$.

    (ii) 'At least one is a girl' selects $\{BG,GB,GG\}$ — $3$ outcomes. Again only $GG$ qualifies, so the probability is $\frac{1}{3}$.

    The answers differ because knowing which child is a girl is stronger information than merely knowing that one of them is.

    ✦ (i) 1/2 (ii) 1/3

13. 133 marksNCERT Exercise 13.1

    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, what is the probability that it will be an easy question given that it is a multiple choice question?

    Hint. Restrict attention to the multiple choice questions only, then find what fraction of those are easy.

    Conditioning on 'multiple choice' means the True/False questions can be discarded entirely.

    The multiple choice questions number $500+400=900$.

    Of these, the easy ones number $500$.

    Therefore the required probability is $\frac{500}{900}=\frac{5}{9}$.

    The total bank size of $1400$ never enters the calculation, because the condition has already removed the other $500$ questions from consideration.

    ✦ P(easy | multiple choice) = 5/9

14. 143 marksNCERT Exercise 13.1

    Given that the two numbers appearing on throwing two dice are different, find the probability of the event 'the sum of numbers on the dice is 4'.

    Hint. Remove the six doubles from the 36 outcomes first, then count which remaining pairs sum to 4.

    Excluding the doubles is the first step, since the condition rules them out.

    Of the $36$ outcomes, the six doubles $(1,1)$ through $(6,6)$ are excluded, leaving $30$ outcomes.

    The pairs summing to $4$ are $(1,3)$, $(2,2)$ and $(3,1)$ — but $(2,2)$ is a double and has been excluded, leaving $2$ outcomes.

    Therefore the probability is $\frac{2}{30}=\frac{1}{15}$.

    Forgetting to drop $(2,2)$ from the numerator is the usual error here.

    ✦ P = 1/15

15. 154 marksNCERT Exercise 13.1

    Consider the experiment of throwing a die. If a multiple of 3 comes up, throw the die again; if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a 3'.

    Hint. Ask when the coin is tossed at all, and whether a 3 can appear on any die in those cases.

    The structure of the experiment settles this without any arithmetic.

    The coin is tossed only when the first throw is not a multiple of $3$, that is when the die shows $1$, $2$, $4$ or $5$.

    In exactly those cases no second die is thrown, and the single die shown is $1$, $2$, $4$ or $5$ — never a $3$.

    So the event 'the coin shows a tail' and the event 'at least one die shows a 3' cannot occur together, making their intersection empty.

    Therefore the conditional probability is $\frac{0}{P(\text{at least one 3})}=0$.

    ✦ P = 0

16. 162 marksNCERT Exercise 13.1

    Choose the correct answer. If $P(A)=\frac{1}{2}$, $P(B)=0$, then $P(A|B)$ is (A) 0 (B) $\frac{1}{2}$ (C) not defined (D) 1

    Hint. Look at the denominator in the definition of conditional probability.

    The definition $P(A|B)=\frac{P(A\cap B)}{P(B)}$ requires dividing by $P(B)$.

    Here $P(B)=0$, so the division is undefined and the conditional probability does not exist.

    This is why every statement of the formula carries the proviso $P(B)\ne 0$. Answering $0$ is the common trap, but a quantity that cannot be computed is not the same as a quantity equal to zero.

    ✦ (C) not defined

17. 172 marksNCERT Exercise 13.1

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

    Hint. Write both conditional probabilities out and compare their denominators.

    Writing each side using the definition makes the conclusion immediate.

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

    Setting them equal gives $\frac{P(A\cap B)}{P(B)}=\frac{P(A\cap B)}{P(A)}$.

    Since the numerators are identical, the denominators must agree, so $P(A)=P(B)$.

    Note this forces only the probabilities to be equal, not the sets themselves, which rules out option (B).

    ✦ (D) P(A) = P(B)

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.