Each of four persons fires one shot at a target Let Ck denot

Each of four persons fires one shot at a target. Let C_k denote the event that the target is hit by person k, k = 1,2,3,4. If C_1, C_2,C3, C_4 are independent and if P(C_1) = P(C_2) = 0.7, P(C_3) = 0.9, and P(C4) = 0.4, compute the probability that (a) all of them hit the target; (b) exactly one hits the target; (c) no one hits the target; (d) at least one hits the target.

Solution

a) P(All of them hit the target) = P(C₁) * P(C₂) * P(C₃) * P(C₄)

                                                = 0.7 * 0.7 * 0.9 * 0.4

                                                = 0.1764

b) P(exactly one hits the target) = P(only 1st person hit) + P(only 2nd person hit) + P(only 3rd person hit) + P(only 4th person hit)

                                                = 0.7 * 0.3 * 0.1 * 0.6 + 0.3 * 0.7 * 0.1 *0.6 + 0.3 * 0.3 * 0.9 * 0.6 + 0.3 * 0.3 * 0.1 * 0.4

                                                = 0.0126 + 0.0126 + 0.0486 + 0.0036

                                                = 0.0774

c) P(no one hits target) = 0.3 * 0.3 * 0.1 * 0.6

                                      = 0.0054

d)P(atleast one hits target) = 1 - P(no one hits target)

                                           = 1 - 0.0054

                                           = 0.9946