I am new into the field of "Probabilites and Statistics" and I try to understand how the following problem can be solved:
There is a target shooting competition. Each participant is allowed to shoot at the target ten times (5 times standing and 5 times lying down). The probability of success at each standing pull is p1 ∈ (0.1) and the probability of success at each standing pull is p2 ∈ (0.1). What is the probability that athlete A will miss the target 3 times?
All can I think of is that it can be solved using Bernoulli distribution but I'm not sure and even if I was I'm kind of confused about how to apply it. The participant could miss the target at any of these 2 kind of shooting (standing and lying down) and I don't know how to combine the success rates in order to obtain the correct result. If anyone can guide me through the process of solving this problem I'll be thankful.
Say you miss $1$ time standing and $2$ times lying down then you have ${5 \choose 1}$ ways to choose the miss standing and ${5 \choose 2}$ ways for the other one.
Then, the probability is $$\left[{5 \choose 1}p_1^4(1-p_1)\right]\left[{5 \choose 2}p_2^3(1-p_2)^2\right]$$
You can easily translate the classical binomial distribution to this case to understand the calculation. If you do the same for the other cases and sum up you are done