If A and B are independent events such that P(A|B) = 1/3 and P(B) = 1/2, then the value of P(A B) is equal to Option 1: 1/3 Option 2: 1/6 Option 3: 1/12 Option 4: 1/2

Answer: 1/6. Solution: Events A and B are independent, with P(A, B) = 1/3 and P(B) = 1/2. For independent events, P(A, B) = P(A). (Translation: If A and B don't affect each other, then the probability of A given B has happened is just the probability of A) Since P(A, B) = 1/3, this means P(A) = 1/3. For independent events: P(A B) = P(A) x P(B) Substituting the values: P(A B) = 1/3 x 1/2 P(A B) = 1/6 Alternatively, using the conditional probability formula: P(A, B) = (P(A B))/(P(B)) Rearranging: P(A B) = P(A, B) x P(B) P(A B) = 1/3 x 1/2 P(A B) = 1/6 Therefore, P(A B) = 1/6.

Algebra

Probability

Medium

core

If A and B are independent events such that $P(A|B) = \frac{1}{3}$ and $P(B) = \frac{1}{2}$ , then the value of $P(A \cap B)$ is equal to

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{1}{12}$

$\frac{1}{2}$

✅ Correct Option: 2

Events $A$ and $B$ are independent, with $P(A|B) = \frac{1}{3}$ and $P(B) = \frac{1}{2}$ .

For independent events, $P(A|B) = P(A)$ .

(Translation: If $A$ and $B$ don't affect each other, then the probability of $A$ given $B$ has happened is just the probability of $A$ )

Since $P(A|B) = \frac{1}{3}$ , this means $P(A) = \frac{1}{3}$ .

For independent events:

$P(A \cap B) = P(A) \times P(B)$

Substituting the values:

$P(A \cap B) = \frac{1}{3} \times \frac{1}{2}$

$P(A \cap B) = \frac{1}{6}$

Alternatively, using the conditional probability formula:

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

Rearranging:

$P(A \cap B) = P(A|B) \times P(B)$

$P(A \cap B) = \frac{1}{3} \times \frac{1}{2}$

$P(A \cap B) = \frac{1}{6}$

Therefore, $P(A \cap B) = \frac{1}{6}$ .

2022: 6 Aug Shift 2

The value of $k$ is

2026: 15 May Shift 1

If E and F are two independent events such that $P(E)=\dfrac{1}{3}, P(F)=\dfrac{3}{5}$ , then which of the following is/are correct ?

A. $P(E \cap F) = \dfrac{1}{5}$

B. $P\left(\dfrac{\overline{E}}{F}\right) = \dfrac{2}{3}$

C. $P\left(\dfrac{E}{\overline{F}}\right) = \dfrac{2}{3}$

D. $P\left(\dfrac{E}{F}\right) = \dfrac{2}{3}$

Choose the correct answer from the options given below:

2025: 13 May Shift 1

A die is thrown three times. If the first throw is a five, the probability of getting 14 as the sum is

2025: 15 May Shift 2

If X is a random variable and a, b are real numbers, then which of the following statements are correct?

(A) $E[aX+b] = a E(X) + b$

(B) $Var (aX + b) = a^2 Var (X) + b$

(D) $Var (X) = E(X^2) - [E(X)]^2$

Choose the correct answer from the options given below:

2026: 18 May Shift 1

A die is thrown and a coin is tossed. The probability that the outcome will be a number greater than 4 or a head or both is

2025: 26 May Shift 2

The random variable X has the following probability distribution

X	0	1	2	3
P(X)	a	a	b	b

such that E(x²) = 2E(x), then the value of b is:

If A and B are independent events such that P(A∣B)=13P(A|B) = \frac{1}{3}P(A∣B)=31​ and P(B)=12P(B) = \frac{1}{2}P(B)=21​, then the value of P(A∩B)P(A \cap B)P(A∩B) is equal to

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If A and B are independent events such that $P(A|B) = \frac{1}{3}$ and $P(B) = \frac{1}{2}$ , then the value of $P(A \cap B)$ is equal to