Question

a missile is fired with 96% accuracy. Each firing is independent of the other. If four missiles are fired what is the probability only the second one misses

Answer 1

Answer: 0.03538944

Step-by-step explanation:

Given, The accuracy of missile = 96%

i.e. The probability that it will hit the shot = 0.96

Probability of missing the shot = 1-0.96= 0.04

Now, each missile is independent of other.

Therefore, the probability only the second one misses = (probability that first will hit the shot) x (probability that second one misses ) x (probability that third will hit the shot)x (probability that fourth will hit the shot)

Thus, the probability only the second one misses is $(0.96)(0.04)(0.96)(0.96)=0.03538944$

Hence, the required probability = 0.03538944