If A and B are two events such that P(A, B) = P(B, A), and A B ≠ then Option 1: P(A) = 1 Option 2: P(A) = P(B) Option 3: P(B) = 1 Option 4: P(A B) = 0

Answer: P(A) = P(B). Solution: Given: P(A, B) = P(B, A) and A B ≠ Since A B ≠, events A and B can occur together, meaning P(A B) > 0. The conditional probability formulas are: P(A, B) = (P(A B))/(P(B)) P(B, A) = (P(A B))/(P(A)) Using the given condition P(A, B) = P(B, A): (P(A B))/(P(B)) = (P(A B))/(P(A)) Cross-multiplying: P(A B) * P(A) = P(A B) * P(B) Since P(A B) > 0, dividing both sides by P(A B): P(A) = P(B) Therefore, P(A) = P(B).

Algebra

Probability

Medium

core

If A and B are two events such that $P(A|B) = P(B|A)$ , and $A \cap B \neq \phi$ then

$P(A) = 1$

$P(A) = P(B)$

$P(B) = 1$

$P(A \cap B) = 0$

✅ Correct Option: 2

Given:

$P(A|B) = P(B|A)$ and $A \cap B \neq \phi$

Since $A \cap B \neq \phi$ , events A and B can occur together, meaning $P(A \cap B) > 0$ .

The conditional probability formulas are:

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

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

Using the given condition $P(A|B) = P(B|A)$ :

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

Cross-multiplying:

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

Since $P(A \cap B) > 0$ , dividing both sides by $P(A \cap B)$ :

$P(A) = P(B)$

Therefore, $P(A) = P(B)$ .

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A and B throw a die alternatively till one of them gets a number more than 4 and wins the game. Then the probability of winning the game by B, if A starts first :

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Urn I contains 6 red balls and 4 black balls and Urn II contains 4 red balls and 6 black balls. One ball is drawn at random from Urn I and placed in Urn II. If one ball is drawn at random from Urn II, then the probability that it is a red ball is :

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In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is

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Two numbers are selected without replacement at random, one at a time from the first six positive integers. Let x denotes the larger of the two numbers.

Match List-I with List-II

List-I	List-II
(A) P(x = 2)	(i) $\frac{4}{15}$
(B) P(x = 3)	(ii) $\frac{1}{15}$
(C) P(x = 4)	(iii) $\frac{2}{15}$
(D) P(x = 5)	(iv) $\frac{1}{5}$

Choose the correct answer from the options given below:

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Let E and F are events associated with an experiment. If $P(E) = 0.4$ , $P(F) = 0.8$ and $P(F|E) = 0.6$ , then $P(E|F)$ is

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If three balls are drawn one by one without replacement from a bag containing 5 white and 4 red balls, then the probability distribution of the number of white balls drawn is

If A and B are two events such that P(A∣B)=P(B∣A)P(A|B) = P(B|A)P(A∣B)=P(B∣A), and A∩B≠ϕA \cap B \neq \phiA∩B=ϕ then

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CUET Mathematics Past Year Questions (Topic-Wise)

If A and B are two events such that $P(A|B) = P(B|A)$ , and $A \cap B \neq \phi$ then