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

Let \(\mathrm{p}\) be the probability of getting even number.
Let random variable \(\mathrm{X} \sim \mathrm{B}(\mathrm{n}, \mathrm{p})\)
Given that \(P(X=3)=2 P(X=2)\)
\(\therefore { }^5 \mathrm{C}_3 \mathrm{p}^3 \mathrm{q}^2=2{ }^5 \mathrm{C}_2 \mathrm{p}^2 \mathrm{q}^3\)
\(\therefore \mathrm{p}=2 \mathrm{q}\)
\(\therefore \mathrm{p}+\mathrm{q}=1 \Rightarrow \mathrm{p}=\frac{2}{3}\) and \(\mathrm{q}=\frac{1}{3}\)
\(\therefore \mathrm{P}(\mathrm{X}=0)={ }^5 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^5=\frac{1}{3^5}\)
\(\therefore \) In 1 set of 5 throws, number of times getting no even number is \(\frac{1}{3^5}\).
\(\therefore \) In 6804 sets of 5 throws, number of times getting no even number is
\(\frac{1}{3^5} \times 6804=28\)