Algebra

Probability

Easy

core

Let E and F be two events such that $P(E) = \frac{1}{3}$ , $P(F) = \frac{1}{4}$ and $P(E \cap F) = \frac{1}{5}$ . Then the value of $P(F|E)$ is equal to

$\frac{4}{5}$

$1$

$\frac{3}{5}$

$\frac{1}{12}$

✅ Correct Option: 3

The question asks: if we already know that event $E$ has happened, what is the probability that event $F$ also happens? This is what conditional probability means.

The formula for conditional probability of $F$ given $E$ is:

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

$P(E) = \dfrac{1}{3}$

$P(F) = \dfrac{1}{4}$

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

Plugging into the formula:

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

$P(F|E) = \dfrac{\frac{1}{5}}{\frac{1}{3}}= \dfrac{1}{5} \times \dfrac{3}{1} = \boxed{\dfrac{3}{5}}$

2024: 16 May Shift 1

The probability of a shooter hitting a target is $\frac34$ . How many minimum number of times must he fire so that the probability of hitting the target at least once is more than $90 \%$ ?

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:

2026: 31 May Shift 2

A die is rolled. If the outcome is an odd number, then the probability that the number is prime, is:

2023: 22 May Shift 3

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 :

2022: 7 Aug Shift 2

In Binomial distribution with parameters $n = 12$ and $p = \frac{1}{3}$ , value of $E(X^2) + E(X)$ is :

2026: 31 May Shift 2

Let X denote the number of hours a person watches television during a randomly selected day. The probability that X can take the value $x_i$ , has the following form, where $k$ is some unknown constant.

$P(X=x_i) = \begin{cases} 0.5 & \text{if } x_i=0 \\ kx_i & \text{if } x_i=1 \\ k(4-x_i) & \text{if } x_i=2 \text{ or } 3 \\ 0 & \text{otherwise} \end{cases}$

The probability that the person watches at most 2 hours of television on a selected day is :

Let E and F be two events such that P(E)=13P(E) = \frac{1}{3}P(E)=31​, P(F)=14P(F) = \frac{1}{4}P(F)=41​ and P(E∩F)=15P(E \cap F) = \frac{1}{5}P(E∩F)=51​. Then the value of P(F∣E)P(F|E)P(F∣E) is equal to

Solution

Related questions:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

Let E and F be two events such that $P(E) = \frac{1}{3}$ , $P(F) = \frac{1}{4}$ and $P(E \cap F) = \frac{1}{5}$ . Then the value of $P(F|E)$ is equal to