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 Option 1: 0.4 Option 2: 0.2 Option 3: 0.3 Option 4: 0.5

Answer: 0.3. Solution: Given: P(E) = 0.4 P(F) = 0.8 P(F, E) = 0.6 We need to find P(E, F), which is the probability of E happening given that F has already happened. To find this, we'll use the conditional probability formula, but first we need to find P(E F). We're given P(F, E) = 0.6. Using the conditional probability formula: P(F, E) = (P(E F))/(P(E)) Rearranging: P(E F) = P(F, E) x P(E) P(E F) = 0.6 x 0.4 P(E F) = 0.24 Now we can find P(E, F) using: P(E, F) = (P(E F))/(P(F)) Substituting the values: P(E, F) = 0.24/0.8 P(E, F) = 0.3 Therefore, P(E, F) = 0.3

Algebra

Probability

Medium

core

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

0.4

0.2

0.3

0.5

✅ Correct Option: 3

Given:

$P(E) = 0.4$

$P(F) = 0.8$

$P(F|E) = 0.6$

We need to find $P(E|F)$ , which is the probability of $E$ happening given that $F$ has already happened.

To find this, we'll use the conditional probability formula, but first we need to find $P(E \cap F)$ .

We're given $P(F|E) = 0.6$ . Using the conditional probability formula:

$P(F|E) = \frac{P(E \cap F)}{P(E)}$

Rearranging:

$P(E \cap F) = P(F|E) \times P(E)$

$P(E \cap F) = 0.6 \times 0.4$

$P(E \cap F) = 0.24$

Now we can find $P(E|F)$ using:

$P(E|F) = \frac{P(E \cap F)}{P(F)}$

Substituting the values:

$P(E|F) = \frac{0.24}{0.8}$

$P(E|F) = 0.3$

Therefore, $P(E|F) = 0.3$

2025: 15 May Shift 1

If $X$ is a normal variate with mean 12 and standard deviation 4, then $P[X \ge 20]$ is:

[Given that: $P[0 \le Z \le 2] = 0.4772]$

2025: 3 June Shift 2

The probabilities of occurrence of two events E and F are 0.25 and 0.50 respectively. The probability of their simultaneous occurrence is 0.14. The probability that neither E nor F occurs is

2025: 13 May Shift 2

Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice.

Match List-I with List-II

List-I	List-II
X	Probability, P(X)
(A) 4	(I) $\frac{1}{6}$
(B) 5	(II) $\frac{5}{36}$
(C) 6	(III) $\frac{1}{12}$
(D) 7	(IV) $\frac{1}{9}$

Choose the correct answer from the options given below:

2025: 21 May Shift 1

If a random variable y follows Poisson's distribution such that $P(X = 2) = 9 P(X = 4) + 90 P(X = 6)$ , then sum of the mean and variance of X is

2025: 3 June Shift 1

60% members of a committee favour a certain proposal and 40% members oppose the proposal. A member is selected and let the random variable X = 0 if he opposes and X = 1 if he is in favour. Then the variance of the random variable X is

2026: 30 May Shift 2

A random variable X has the following probability distribution:

X	0	1	2	3	4	5	6
P(X)	z	2z	2z	3z	$z^2$	$2z^2$	$7z^2+z$

Which of the following are TRUE?

(A) P(X < 3) = 0.8

(B) P(X > 4) = 0.2

(D) $P(3\le X<5) = 0.31$

Choose the correct answer from the options given below:

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

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CUET Mathematics Past Year Questions (Topic-Wise)

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