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

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 $90 \%$ ?

A random variable X has the following probability distribution:

Which of the following are TRUE?

(A) P(X < 3) = 0.8

(B) P(X > 4) = 0.2

(C) The value of z is 0.1

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

Choose the correct answer from the options given below:

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

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 :

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

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 :