An irregular six faced die is thrown and the probability that, in 5 throws it will give 3 even numbers is…
An irregular six faced die is thrown and the probability that, in 5 throws it will give 3 even numbers is twice the probability that it will give 2 even numbers. The number of times, in 6804 sets of 5 throws, you expect to give no even number is
$18$
$28$
$27$
$19$
Solution
Let $\mathrm{p}$ be the probability of getting even number.
Let random variable $\mathrm{X} \sim \mathrm{B}(\mathrm{n}, \mathrm{p})$
Given that $P(X=3)=2 P(X=2)$
$\therefore \quad{ }^5 \mathrm{C}_3 \mathrm{p}^3 \mathrm{q}^2=2{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3$
$\therefore \quad \mathrm{p}=2 \mathrm{q}$
$\therefore \quad \mathrm{p}+\mathrm{q}=1 \Rightarrow \mathrm{p}=\frac{2}{3}$ and $\mathrm{q}=\frac{1}{3}$
$\therefore \quad \mathrm{P}(\mathrm{X}=0)={ }^5 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^5=\frac{1}{3^5}$
$\therefore \quad$ In 1 set of 5 throws, number of times getting no even number is $\frac{1}{3^5}$.
$\therefore \quad$ In 6804 sets of 5 throws, number of times getting no even number is
$\frac{1}{3^5} \times 6804=28$