An irregular 6 faced dice is such that the probability that it gives 3 even numbers in 5 throws is twice the probability that it gives 2 even numbers in 5 throws. How many sets of exactly 5 trials can be expected to give no even number out of 2500 sets?
(1) Nearly 5 (2) Nearly 10 (3) Nearly 15 (4) Nearly 20
Let the probability of an even number be $p$. Thus the probability of an odd number is $1-p$.
The probability of $3$ even in $5$ throws is $\binom{5}{3}p^3(1-p)^2$. The probability of $2$ even in $5$ throws is $\binom{5}{2}p^2(1-p)^3$.
We are told that $\binom{5}{3}p^3(1-p)^2=2\binom{5}{2}p^2(1-p)^3$. Solve for $p$. There are the solutions $p=0$ and $p=1$, which we are presumably expected to discard, although there is no mathematical reason to do so. In addition, we have the solution $p=\frac{2}{3}$. So the probability that a single toss results in a number that is not even is $\frac{1}{3}$.
The probability of no even in $5$ trials is therefore $\frac{1}{3^5}$.
Now in $2500$ sets of $5$ trials, the expected number of times we have no even is $2500\cdot\frac{1}{3^5}$.
