Recent content by Rojito

Thanks Soroban, I was hoping for some help on the rest of the question, which I'm totally stuck on.

Would I be correct in assuming that the chance of a mishit is .25 \therefore 3 mishits = 1/4.1/4.1/4=1/64 The next attempt (4th shot) = 3/4.1/64=3/256

(a) If the probability of hitting a target is ¾ (0.75), find the probability of hitting the target first on the fourth shot. More generally, if the probability of hitting the target is P, what is the probability of hitting the target for the first time on the nth shot? (b) A computer system in...