There are three shooters ( i= 1,2,3) and they hit the target with probabilities

P(A1) = 1/2, P(B1) = 2/3, P(A3) = 3/4. All men shoot at a same time and

results are independent.

- What is the probability that there will be two hits?

P(A1) intersection P(B1) intersection P(A3) =

or 1/2 + 2/3 + 3/4 / 3 = 0.32

or 1/2 * 2/3 * 3/4 = 0.25

or 1/2 * 2/3 * 3/4 / 2 = 0.125

or 1/2 * 2/3 * 3/4 ^2 = 0.0625

- What is the probability that one shooter missed?

P(A1) * P(B1) * P(A3) / union ...

1/2 * 2/3 * 3/4 / 3 = 0.083

I don't know. Can you please help?