Question

A die is thrown $2n$ times. The probability that the number greater than 4 appears at least once in $2n$ throws is

• A. ${\left(\frac{1}{3}\right)}^{2n}$
• B. $1-{\left(\frac{1}{3}\right)}^{2n}$
• C. $\frac{3^{2n}-2^{2n}}{3^{2n}}$
• D. None of these

Answer 1

The correct option is C $\frac{3^{2n}-2^{2n}}{3^{2n}}$

Probability of success, $p=P(X>4)=P(5$ or $6)=\frac{2}{6}=\frac{1}{3}$ $\therefore q=1-\frac{1}{3}=\frac{2}{3}$
Now using binomial distribution, $P{(X=r)}^{2n}{C}_r{\left(\frac{1}{3}\right)}^r{\left(\frac{2}{3}\right)}^{2n-r}$
$P(1$ or $2$ or $\mathrm{3...}$ or $2n)=P(1)+P(2)+...+p(2n)=1-P(0)$
$=1{-}^{2n}{C}_0{\left(\frac{1}{3}\right)}^0{\left(\frac{2}{3}\right)}^{2n}$ $=(1-\frac{2^{2n}}{3^{2n}})=\frac{3^{2n}-2^{2n}}{3^{2n}}$