Percentage chace of things happening I have 5 mystery boxes with percentage odds of containing a prize. 1 has a 55% chance of containing a prize, the other 4 have a 50% chance of containing a prize. What is the percentage chance of just 3 having a prize in?
How do i work this out myself as i will also need to find out the percentage chance of 2, 1 and none having a prize?
 A: We can have one of $55\%$ chance and other $2$ from $50\%$ chance out of $4$  whose probability is
$$\frac{55}{100}\cdot1\cdot\binom 42\left(\frac12\right)^{4-2}\left(1-\frac12\right)^2$$
or all $3$ from $50\%$ chance out of $4$ whose probability is 
$$\left(1-\frac{55}{100}\right) \binom 43\left(\frac12\right)^{4-3}\left(1-\frac12\right)^3$$
A: We want to find the probability that exactly $3$ have a prize. This can happen in two disjoint ways: (i) the first box has a prize and exactly $2$ of the others have a prize or (ii) the first box has no prize, and exactly $3$ of the others have a prize.
We first calculate the probability of (i). The probability of a prize in the first box is $0.55$. Given this has happened, the probability of exactly $2$ prizes in the remaining $4$ boxes is the probability of $2$ heads in $4$ tosses of a fair coin, which is $\binom{4}{2}\left(\frac{1}{2}\right)^4$. So the probability of $(i)$ is $(0.55)\binom{4}{2}\left(\frac{1}{2}\right)^4$.
The probability of (ii) can be calculated in the same way. the probability of no prize in te first box is $0.45$, and the probability of $3$ prizes in the remaining boxes is $\binom{4}{2}\left(\frac{1}{2}\right)^4$, for a probability of 
$(0.45)\binom{4}{3}\left(\frac{1}{2}\right)^4$.
Thus our required probability is
$$(0.55)\binom{4}{2}\left(\frac{1}{2}\right)^4+(0.45)\binom{4}{3}\left(\frac{1}{2}\right)^4.$$
Calculate, and convert to a percentage. Note that $\binom{4}{2}=6$ and $\binom{4}{3}=4$.
The same sort of analysis will take care of the remaining problems mentioned, though $0$ prizes does not require the full analysis. 
Remark: We have assumed that the events prize/no prize in the various boxes are independent. If we do not have independence (and we need not), then we need additional information to solve the problem. 
