# The probability of hitting a target for a given shooter is on average 80%. The shooter fired 6 shots

**The probability of hitting a target for a given shooter is on average 80%. The shooter fired 6 shots at the target. Find the probability that the target was hit: a) five times; b) at least five times; c) no more than five times.**

Let’s designate the events of hitting the target during the first, second … sixth shot as A1, A2, A3, A4, A5, A6. The events of miss during shots are denoted as B1, B2, B3, B4, B5, B6.

a) Find the probability that the shooter has hit five times out of six shots.

P1 = (B1 * A2 * A3 * A4 * A5 * A6) + (A1 * B2 * A3 * A4 * A5 * A6) + (A1 * A2 * B3 * A4 * A5 * A6) + (A1 * A2 * A3 * B4 * A5 * A6) + (A1 * A2 * A3 * A4 * B5 * A6) + (A1 * A2 * A3 * A4 * A5 * B6) = 0.2 * 0.8 * 0.8 * 0, 8 * 0.8 * 0.8 + 0.8 * 0.2 * 0.8 * 0.8 * 0.8 * 0.8 + 0.8 * 0.8 * 0.2 * 0.8 * 0.8 * 0.8 + 0.8 * 0.8 * 0.8 * 0.2 * 0.8 * 0.8 + 0.8 * 0.8 * 0.8 * 0.8 * 0, 2 * 0.8 + 0.8 * 0.8 * 0.8 * 0.8 * 0.8 * 0.2 = 0.393216

b) The probability that the shooter has hit at least five times will be equal to the sum of the probabilities that the shooter has hit five times which we have already found and the probability that the shooter has hit the target all six times:

P2 = P1 + (A1 * A2 * A3 * A4 * A5 * A6) = 0.393216 + 0.262144 = 0.65536

c) The probability that the shooter has hit no more than five times will be equal to:

P3 = 1 – (A1 * A2 * A3 * A4 * A5 * A6) = 1 – 0.262144 = 0.737856