Can you predict the probability. .??

A probability of having $5$ scores out of $10$ is simple, without considering the order, it would be $P(score)^{5} * P(fails)^{5} = (\frac{1}{2})^{5} * (\frac{1}{2})^{5} = 2^{-10}$ .

So, at this point, the only things that one need to find out is the amount of orders to score $4$ out of $9$ , which is $\frac{9!}{4!5!}$ , as the last shot will always scored, .

Finally, the probability of hitting the target $5$ th time on the $10$ th trial would be $2^{-10} * \frac{9!}{4!5!}$ which is easily calculated as $0.123046875$ .

• Let H be hitting and N be not hitting:
• In the first nine trials we must have four H and five N, we can accomodate this in $\frac{9!}{4! \times 5!} = 126.$
• But H and A have $\frac{1}{2}$ of occurring so we must divide 126 by $2^{9}$ and then multiplied by the chance of the 10th trial be a H, which is $\frac{1}{2}$ .
• the probability is then ( \frac{126}{2^{10}} = 0.1230468