Let A and B be two independent events such that P(A)=k, P(B)=2k and probability of occurrence of exactly one of them is 5/9. Then the value(s) of k is/are: Option 1: k=1/3 Option 2: k=5/12 Option 3: k=1/3,5/12 Option 4: k=2/3,7/12

Answer: k=1/3,5/12. Solution: For independent events, P(exactly one) = P(A)+P(B)-2P(A)P(B) = 3k-4k^2. Setting 3k-4k^2 = 5/9 gives 36k^2-27k+5 = 0, so k = 1/3 or k = 5/12. Both satisfy 2k ≤ 1, so both are admissible.

Algebra

Probability

Medium

core

Let A and B be two independent events such that $P(A)=k$ , $P(B)=2k$ and probability of occurrence of exactly one of them is $\dfrac{5}{9}$ . Then the value(s) of $k$ is/are:

$k=\dfrac{1}{3}$

$k=\dfrac{5}{12}$

$k=\dfrac{1}{3},\dfrac{5}{12}$

$k=\dfrac{2}{3},\dfrac{7}{12}$

✅ Correct Option: 3

For independent events, $P(\text{exactly one}) = P(A)+P(B)-2P(A)P(B) = 3k-4k^2$ .

Setting $3k-4k^2 = \dfrac{5}{9}$ gives $36k^2-27k+5 = 0$ , so $k = \dfrac{1}{3}$ or $k = \dfrac{5}{12}$ .

Both satisfy $2k \le 1$ , so both are admissible.

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(B) $E(X^2) = 4+7\alpha$

(D) $\alpha = \frac{1}{5}$

Choose the correct answer from the options given below:

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Let A and B be two independent events such that P(A)=kP(A)=kP(A)=k, P(B)=2kP(B)=2kP(B)=2k and probability of occurrence of exactly one of them is 59\dfrac{5}{9}95​. Then the value(s) of kkk is/are:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let A and B be two independent events such that $P(A)=k$ , $P(B)=2k$ and probability of occurrence of exactly one of them is $\dfrac{5}{9}$ . Then the value(s) of $k$ is/are: