Exercise 13.2 — Probability

1. 12 marksNCERT Exercise 13.2

   If $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$, find $P(A\cap B)$ if A and B are independent events.

   Hint. For independent events the intersection is simply the product.

   Independence is defined by exactly the multiplication property, so no further work is needed.

   $P(A\cap B)=P(A)\cdot P(B)=\frac{3}{5}\times \frac{1}{5}=\frac{3}{25}$

   This multiplication is valid only because independence was given; for general events the intersection cannot be recovered from the two separate probabilities.

   ✦ P(A intersect B) = 3/25

2. 23 marksNCERT Exercise 13.2

   Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

   Hint. Without replacement, the second draw sees one fewer card in total and one fewer black card.

   Drawing without replacement makes the second probability conditional on the first, so the multiplication rule is used rather than a plain product.

   There are $26$ black cards. The first draw is black with probability $\frac{26}{52}$.

   Given that, $25$ black cards remain among $51$, so the second is black with probability $\frac{25}{51}$.

   Hence $P=\frac{26}{52}\times \frac{25}{51}=\frac{1}{2}\times \frac{25}{51}=\frac{25}{102}$.

   ✦ P = 25/102

3. 33 marksNCERT Exercise 13.2

   A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all three oranges are good, the box is approved for sale, otherwise it is rejected. Find the probability that a box containing 15 oranges, of which 12 are good and 3 are bad, will be approved for sale.

   Hint. Multiply three conditional probabilities, reducing both the good count and the total each time.

   Approval requires all three drawn oranges to be good, and each draw changes the composition of the box.

   The first orange is good with probability $\frac{12}{15}$.

   Given that, $11$ good remain out of $14$, so the second is good with probability $\frac{11}{14}$.

   Given both, $10$ good remain out of $13$, giving $\frac{10}{13}$.

   Multiplying, $P=\frac{12}{15}\times \frac{11}{14}\times \frac{10}{13}=\frac{1320}{2730}=\frac{44}{91}$.

   ✦ P = 44/91

4. 43 marksNCERT Exercise 13.2

   A fair coin and an unbiased die are tossed. Let A be the event 'head appears on the coin' and B be the event '3 on the die'. Check whether A and B are independent events or not.

   Hint. Compute P(A), P(B) and P(A intersect B) separately, then test whether the product rule holds.

   Testing independence always means comparing the joint probability against the product of the separate ones.

   $P(A)=\frac{1}{2}$ and $P(B)=\frac{1}{6}$.

   The combined sample space has $2\times 6=12$ equally likely outcomes, of which only (head, 3) satisfies both, so $P(A\cap B)=\frac{1}{12}$.

   Since $P(A)\cdot P(B)=\frac{1}{2}\times \frac{1}{6}=\frac{1}{12}=P(A\cap B)$, the condition holds.

   This is unsurprising, because the coin and the die are physically unrelated.

   ✦ A and B are independent

5. 53 marksNCERT Exercise 13.2

   A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event 'the number is even' and B be the event 'the number is red'. Are A and B independent?

   Hint. Here one die carries both attributes, so the events can interact even though they describe different features.

   Both events live on the same single throw, so there is no reason to expect independence in advance.

   $A=\{2,4,6\}$ gives $P(A)=\frac{1}{2}$, and $B=\{1,2,3\}$ gives $P(B)=\frac{1}{2}$.

   The only even red number is $2$, so $A\cap B=\{2\}$ and $P(A\cap B)=\frac{1}{6}$.

   But $P(A)\cdot P(B)=\frac{1}{4}$, and $\frac{1}{6}\ne \frac{1}{4}$.

   Because the product rule fails, the events are not independent.

   ✦ A and B are NOT independent

6. 63 marksNCERT Exercise 13.2

   Let E and F be events with $P(E)=\frac{3}{5}$, $P(F)=\frac{3}{10}$ and $P(E\cap F)=\frac{1}{5}$. Are E and F independent?

   Hint. Put both quantities over a common denominator to compare them cleanly.

   A common denominator makes the comparison unambiguous.

   $P(E)\cdot P(F)=\frac{3}{5}\times \frac{3}{10}=\frac{9}{50}$.

   Meanwhile $P(E\cap F)=\frac{1}{5}=\frac{10}{50}$.

   Since $\frac{10}{50}\ne \frac{9}{50}$, the defining product rule fails.

   The two values are close but not equal, which is exactly why the comparison must be done exactly rather than by decimal approximation.

   ✦ E and F are NOT independent

7. 74 marksNCERT Exercise 13.2

   Given that the events A and B are such that $P(A)=\frac{1}{2}$, $P(A\cup B)=\frac{3}{5}$ and $P(B)=p$. Find $p$ if they are (i) mutually exclusive (ii) independent.

   Hint. The addition rule takes a different form in each case, because the intersection differs.

   The addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ applies in both parts; only the intersection term changes.

   (i) Mutually exclusive means $P(A\cap B)=0$, so $\frac{3}{5}=\frac{1}{2}+p$, giving $p=\frac{3}{5}-\frac{1}{2}=\frac{1}{10}$.

   (ii) Independent means $P(A\cap B)=P(A)P(B)=\frac{p}{2}$, so $\frac{3}{5}=\frac{1}{2}+p-\frac{p}{2}=\frac{1}{2}+\frac{p}{2}$.

   Hence $\frac{p}{2}=\frac{1}{10}$ and $p=\frac{1}{5}$.

   The two answers differ because mutual exclusivity removes the overlap entirely, while independence merely fixes its size.

   ✦ (i) p = 1/10 (ii) p = 1/5

8. 83 marksNCERT Exercise 13.2

   Let A and B be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find (i) $P(A\cap B)$ (ii) $P(A\cup B)$ (iii) $P(A|B)$ (iv) $P(B|A)$.

   Hint. Once the intersection is known, note what independence does to the two conditional probabilities.

   Independence simplifies every part, and the last two parts require no computation at all.

   (i) $P(A\cap B)=0.3\times 0.4=0.12$

   (ii) $P(A\cup B)=0.3+0.4-0.12=0.58$

   (iii) $P(A|B)=\frac{0.12}{0.4}=0.3=P(A)$

   (iv) $P(B|A)=\frac{0.12}{0.3}=0.4=P(B)$

   Parts (iii) and (iv) illustrate the meaning of independence: conditioning on one event leaves the other's probability unchanged.

   The conditional probabilities equal the unconditional ones precisely because the events are independent.

   ✦ (i) 0.12 (ii) 0.58 (iii) 0.3 (iv) 0.4

9. 93 marksNCERT Exercise 13.2

   If A and B are two events such that $P(A)=\frac{1}{4}$, $P(B)=\frac{1}{2}$ and $P(A\cap B)=\frac{1}{8}$, find $P$(not A and not B).

   Hint. Recognise 'not A and not B' as the complement of the union, via De Morgan's law.

   De Morgan's law converts the required event into a complement of a union, which the addition rule can handle.

   $P(\text{not }A\text{ and not }B)=P(A^{\prime }\cap B^{\prime })=P((A\cup B{)}^{\prime })=1-P(A\cup B)$.

   First, $P(A\cup B)=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2}{8}+\frac{4}{8}-\frac{1}{8}=\frac{5}{8}$.

   Therefore $P(A^{\prime }\cap B^{\prime })=1-\frac{5}{8}=\frac{3}{8}$.

   The complement route is used here because the union is far easier to compute than the joint complement directly.

   ✦ P(not A and not B) = 3/8

10. 104 marksNCERT Exercise 13.2

    Events A and B are such that $P(A)=\frac{1}{2}$, $P(B)=\frac{7}{12}$ and $P$(not A or not B)$=\frac{1}{4}$. State whether A and B are independent.

    Hint. Convert 'not A or not B' into a statement about the intersection using De Morgan's law.

    De Morgan's law turns the given information into a value for the intersection.

    Since $A^{\prime }\cup B^{\prime }=(A\cap B{)}^{\prime }$, we have $P(A\cap B)=1-\frac{1}{4}=\frac{3}{4}$.

    For independence we would need $P(A)\cdot P(B)=\frac{1}{2}\times \frac{7}{12}=\frac{7}{24}$.

    Since $\frac{3}{4}=\frac{18}{24}\ne \frac{7}{24}$, the events are not independent.

    A note on the printed data. The value $P(A\cap B)=\frac{3}{4}$ that this question forces is larger than $P(A)=\frac{1}{2}$, which is impossible, since an intersection can never be more probable than either event containing it. The question as printed is therefore internally inconsistent. The expected answer remains 'not independent', which is what the product-rule test gives.

    ✦ A and B are NOT independent (note: the printed data is internally inconsistent, since it forces P(A intersect B) = 3/4 > P(A) = 1/2)

11. 113 marksNCERT Exercise 13.2

    Given two independent events A and B such that $P(A)=0.3$, $P(B)=0.6$, find (i) $P$(A and B) (ii) $P$(A and not B) (iii) $P$(A or B) (iv) $P$(neither A nor B).

    Hint. If A and B are independent, then so are A and B', and A' and B'.

    A useful consequence of independence is that it survives complementation: if $A$ and $B$ are independent, so are $A$ and $B^{\prime }$, and $A^{\prime }$ and $B^{\prime }$.

    (i) $P(A\cap B)=0.3\times 0.6=0.18$

    (ii) $P(A\cap B^{\prime })=0.3\times 0.4=0.12$

    (iii) $P(A\cup B)=0.3+0.6-0.18=0.72$

    (iv) $P(A^{\prime }\cap B^{\prime })=0.7\times 0.4=0.28$

    As a check, (iii) and (iv) sum to $1$, as they must, being complementary events.

    ✦ (i) 0.18 (ii) 0.12 (iii) 0.72 (iv) 0.28

12. 123 marksNCERT Exercise 13.2

    A die is tossed thrice. Find the probability of getting an odd number at least once.

    Hint. 'At least once' is easiest through its complement: not even once.

    Attacking 'at least once' directly would need three separate cases, so the complement is far quicker.

    The complement of 'an odd number at least once' is 'no odd number at all', that is an even number on all three throws.

    Each throw is even with probability $\frac{3}{6}=\frac{1}{2}$, and the throws are independent, so $P(\text{all even})={\left(\frac{1}{2}\right)}^3=\frac{1}{8}$.

    Therefore $P(\text{at least one odd})=1-\frac{1}{8}=\frac{7}{8}$.

    ✦ P = 7/8

13. 134 marksNCERT Exercise 13.2

    Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that (i) both balls are red (ii) first ball is black and second is red (iii) one of them is black and the other is red.

    Hint. With replacement, the two draws are independent and the probabilities stay the same each time.

    Replacement restores the box to its original state, so the two draws are independent with unchanging probabilities.

    There are $18$ balls, giving $P(\text{red})=\frac{8}{18}=\frac{4}{9}$ and $P(\text{black})=\frac{10}{18}=\frac{5}{9}$.

    (i) $P(\text{both red})=\frac{4}{9}\times \frac{4}{9}=\frac{16}{81}$

    (ii) $P(\text{black then red})=\frac{5}{9}\times \frac{4}{9}=\frac{20}{81}$

    (iii) 'One black and one red' allows either order, so it is $2\times \frac{5}{9}\times \frac{4}{9}=\frac{40}{81}$.

    The factor of $2$ in part (iii) is what distinguishes it from part (ii), where the order was specified.

    ✦ (i) 16/81 (ii) 20/81 (iii) 40/81

14. 144 marksNCERT Exercise 13.2

    Probability of solving a specific problem independently by A and B are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved (ii) exactly one of them solves the problem.

    Hint. 'Solved' means at least one succeeds, so use the complement. 'Exactly one' means one succeeds while the other fails.

    The two parts need different treatments, since 'solved' and 'exactly one solved' are different events.

    (i) The problem is unsolved only if both fail, which happens with probability $\frac{1}{2}\times \frac{2}{3}=\frac{1}{3}$. Therefore $P(\text{solved})=1-\frac{1}{3}=\frac{2}{3}$.

    (ii) Exactly one solves it in two mutually exclusive ways: A succeeds and B fails, or A fails and B succeeds.

    $P=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$

    Note that (ii) is smaller than (i), since 'solved' also includes the case where both succeed.

    ✦ (i) 2/3 (ii) 1/2

15. 154 marksNCERT Exercise 13.2

    One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent? (i) E : the card drawn is a spade, F : the card drawn is an ace (ii) E : the card drawn is black, F : the card drawn is a king (iii) E : the card drawn is a king or queen, F : the card drawn is a queen or jack.

    Hint. For each pair compute P(E), P(F) and P(E intersect F), then test the product rule.

    Each part is a separate application of the same product-rule test.

    (i) $P(E)=\frac{13}{52}=\frac{1}{4}$, $P(F)=\frac{4}{52}=\frac{1}{13}$, and the ace of spades gives $P(E\cap F)=\frac{1}{52}$. Since $\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}$, they are independent.

    (ii) $P(E)=\frac{1}{2}$, $P(F)=\frac{1}{13}$, and the two black kings give $P(E\cap F)=\frac{2}{52}=\frac{1}{26}$. Since $\frac{1}{2}\times \frac{1}{13}=\frac{1}{26}$, they are independent.

    (iii) $P(E)=\frac{8}{52}=\frac{2}{13}$ and $P(F)=\frac{8}{52}=\frac{2}{13}$, while the overlap is the four queens, giving $P(E\cap F)=\frac{4}{52}=\frac{1}{13}$. Here $\frac{2}{13}\times \frac{2}{13}=\frac{4}{169}$, but $\frac{1}{13}=\frac{13}{169}$, so they are not independent.

    ✦ (i) independent (ii) independent (iii) NOT independent

16. 164 marksNCERT Exercise 13.2

    In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both. A student is selected at random. (a) Find the probability that she reads neither Hindi nor English newspapers. (b) If she reads Hindi newspaper, find the probability that she reads English newspaper. (c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

    Hint. Part (a) uses the complement of the union; parts (b) and (c) are conditional probabilities with different denominators.

    Let $H$ and $E$ denote reading the Hindi and English newspapers, so $P(H)=0.6$, $P(E)=0.4$ and $P(H\cap E)=0.2$.

    (a) $P(H\cup E)=0.6+0.4-0.2=0.8$, so $P(\text{neither})=1-0.8=0.2$.

    (b) $P(E|H)=\frac{0.2}{0.6}=\frac{1}{3}$

    (c) $P(H|E)=\frac{0.2}{0.4}=\frac{1}{2}$

    Parts (b) and (c) share a numerator but differ in denominator, which is why the answers differ — another reminder that conditional probability is not symmetric.

    ✦ (a) 0.2 (b) 1/3 (c) 1/2

17. 172 marksNCERT Exercise 13.2

    Choose the correct answer. The probability of obtaining an even prime number on each die, when a pair of dice is rolled, is (A) 0 (B) $\frac{1}{3}$ (C) $\frac{1}{12}$ (D) $\frac{1}{36}$

    Hint. Identify which numbers on a die are both even and prime.

    The key is to notice how few numbers satisfy 'even prime'.

    The only even prime number is $2$, since every other even number is divisible by $2$ and therefore not prime.

    So each die must show a $2$, which has probability $\frac{1}{6}$ per die.

    The dice are independent, so the probability of both showing $2$ is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$.

    ✦ (D) 1/36

18. 182 marksNCERT Exercise 13.2

    Choose the correct answer. Two events A and B will be independent if (A) A and B are mutually exclusive (B) $P(A^{\prime }{B}^{\prime })=[1-P(A)][1-P(B)]$ (C) $P(A)=P(B)$ (D) $P(A)+P(B)=1$

    Hint. Test each option against the definition; recall that independence is preserved under complementation.

    Independence means $P(A\cap B)=P(A)P(B)$, and the correct option must be equivalent to it.

    Option (B) states $P(A^{\prime }\cap B^{\prime })=P(A^{\prime })P(B^{\prime })$, which is precisely the independence condition applied to the complements. Since $A,B$ are independent exactly when $A^{\prime },B^{\prime }$ are, this is equivalent to the definition and is correct.

    Option (A) is wrong because mutually exclusive events with non-zero probabilities are actually dependent — if one occurs the other cannot.

    Options (C) and (D) impose numerical relations between the probabilities that say nothing about how the events interact.

    ✦ (B) P(A'B') = [1 - P(A)][1 - 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.