Question

A die is thrown n times. A random variable X denotes the number of times, the number on the dice is greater than 4 and P(X = 1) = 2P(X = 2). The value of n is :

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The Correct Option is B

The problem involves determining the number of times (n) a die is thrown such that the probability P(X = 1) is double the probability P(X = 2), where X is the number of times a number greater than 4 is rolled.

When a die is tossed, numbers greater than 4 are 5 and 6. The probability of rolling a number greater than 4, denoted as p, is:

$\frac{2}{6}=\frac{1}{3}$

Probability of not rolling a number greater than 4, denoted as q, is:

$q=1-\frac{1}{3}=\frac{2}{3}$

Using the binomial distribution, the probability P(X = r) is given by:

$P(X=r)=\left(\frac{n}{r}\right)p^r{q}^{(}$

Given $P(X=1)=2P(X=2)$, we have:

$\left(\frac{n}{1}\right){\frac{1}{3}}^1{\frac{2}{3}}^n=2\left(\frac{n}{2}\right){\frac{1}{3}}^2{\frac{2}{3}}^n$

Simplifying further,

${\frac{2}{3}}^{\mathrm{n}}=2\frac{n}{2}\frac{1}{3}\frac{2}{3}$

Rearranging, we get:

$\frac{2}{3}{\frac{2}{3}}^n$

We equate coefficients:

$3n=\frac{n}{2}2$

Solving it:

$n=3$

Therefore, the value of n is 3.