Question

An irregular six faced die is thrown and the expectation that in $10$ throws it will give $5$ even numbers is twice the expectation that it will give $4$ even numbers. The number of times in $10000$ sets of $10$ throws you expect to give no even numbers is

• A. $6$
• B. $1$
• C. $2$
• D. $9$

Answer 1

The correct option is B $1$

Let the random variable $X$ denote the number of even numbers.
$P$(getting x even numbers)
$=P(X=x)$ ${}^n{C}_x{p}^x{q}^{10-x},x=0,1,2,...,n$
Given $n=10$
$\therefore P(X=x){=}^{10}{C}_x{p}^x{q}^{10-x},x=0,1,2,...10$
$P$(getting $5$ even numbers) $=2P$(getting $4$ even numbers)
$P(X=5)=2P(X=4){=}^{10}{C}_5{p}^5{q}^5$

$=2\left({}^{10}{C}_4{p}^4{q}^6\right)$

$\frac{p}{4}=\frac{q}{3}\Rightarrow 3p=5q=5(1-p)$

$\Rightarrow 8p=5\Rightarrow p=\frac{5}{8}$

$\therefore q=1-p=1-\frac{5}{8}=\frac{3}{8}$
$\therefore P$(getting x even numbers)
${}^{10}{C}_x{\left(\frac{5}{8}\right)}^x{\left(\frac{3}{8}\right)}^{10-x},x=0,1,2,...n$
$\therefore$ the required number of times that in $10000$ sets of $10$ throws each we get no even number.
$=1000\times P(X=0)=1000{\times }^{10}{C}_0{\left(\frac{5}{8}\right)}^0{\left(\frac{3}{8}\right)}^{10}=1$ (approx)