Let A and B be two events such that P(A B) = 0 then P(A|(A B)) is equal to: Option 1: (P(A))/(P(B)) Option 2: (P(A))/(P(A) - P(B)) Option 3: (P(A))/(P(A) + P(B)) Option 4: (P(A))/(P(A).P(B))

Answer: (P(A))/(P(A) + P(B)). Solution: Since A A B, the numerator P(A(A B)) = P(A). As P(A B)=0, P(A B) = P(A)+P(B). Hence the conditional probability is P(A)/(P(A)+P(B)).

Algebra

Probability

Medium

core

Let A and B be two events such that $P(A \cap B) = 0$ then $P(A|(A \cup B))$ is equal to:

$\dfrac{P(A)}{P(B)}$

$\dfrac{P(A)}{P(A) - P(B)}$

$\dfrac{P(A)}{P(A) + P(B)}$

$\dfrac{P(A)}{P(A).P(B)}$

✅ Correct Option: 3

Since $A \subseteq A\cup B$ , the numerator $P(A\cap(A\cup B)) = P(A)$ .

As $P(A\cap B)=0$ , $P(A\cup B) = P(A)+P(B)$ .

Hence the conditional probability is $P(A)/(P(A)+P(B))$ .

2025: 14 May Shift 1

A black and a red die are rolled simultaneously. The probability of obtaining a sum greater than 9, given that the black resulted in a 5 is

2025: 19 May Shift 1

If X is a normal variate with mean 16 and standard deviation 4, then the value of standard normal variate Z corresponding to X = 17 is:

2026: 19 May Shift 2

A beg contains 4 white and 6 red balls. Four balls are drawn at random, then match the probability of the numbers of white balls drawn. Where P(E) denotes the probability of an event E.

Then Match List-I with List-II

LIST-I		LIST-II
A.	P (no white ball)	I.	$\frac{8}{21}$
B.	P (all white balls)	II.	$\frac{1}{14}$
C.	P (one white ball)	III.	$\frac{6}{14}$
D.	P (two white balls)	IV.	$\frac{1}{210}$

Choose the correct answer from the options given below:

2025: 3 June Shift 1

If the difference between mean and variance of a Binomial distribution is 1 and the difference of their squares is 5, then the probability of success is

2026: 22 May Shift 2

Five bad apples are mixed with 15 good apples accidently. Let the random variable X be the number of bad apples in a draw of two apples, then mean of X is

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

Let A and B be two events such that P(A∩B)=0P(A \cap B) = 0P(A∩B)=0 then P(A∣(A∪B))P(A|(A \cup B))P(A∣(A∪B)) is equal to:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

Let A and B be two events such that $P(A \cap B) = 0$ then $P(A|(A \cup B))$ is equal to: