Algebra

Probability

Medium

applied

A random variable 'X' denotes the number of sixes obtained in three throws of a die. Then, the mean of the distribution is:-

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{2}{3}$

✅ Correct Option: 2

A random variable $X$ denotes the number of sixes obtained in three throws of a die.

This is a binomial distribution problem with:

Fixed number of trials (3 throws)
Two possible outcomes for each throw (six or not six)
Constant probability for each throw
Independent throws

For a binomial distribution, the key values are:

$n = 3$ (number of trials)

$p = \dfrac{1}{6}$ (probability of getting a six in one throw, since a die has 6 faces)

For a binomial distribution, the mean is given by:

Mean $= n \times p$

Mean $= 3 \times \dfrac{1}{6}$

Mean $= \dfrac{3}{6}$

Mean $= \dfrac{1}{2}$

Therefore, the mean of the distribution is $\dfrac{1}{2}$ .

2026: 22 May Shift 1

Match List-I with List-II

List-I	List-II
(A) For a random variable X, $E(X)=3$ and $E(X^2)=11$ , then $Var(X)$ is	(I) 25
(B) For a random variable X, $E(X)=3$ and $E(X^2)=13$ , then $Var\left(\frac{X}{2}\right)$ is	(II) 2
(C) In a binomial distribution, if $n=20, q=0.75$ , then its mean is	(III) 1
(D) If in a binomial distribution, mean is 5 and variance is 4, then the number of trials is	(IV) 5

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Solution

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