Q. An anti aircraft gun fires at a moving enemy plane with 4 successive shots. Probabilities of the shots S1, S2, S3 and S4 hitting the plane are 0.4, 0.3, 0.2, 0.1 respectively.
(a) What is the probability that the gun is able to hit the plane?
Correct Answer is: 0.6976
(b) What is the conditional probability that at least three shots are required to be shot?
Correct Answer is: 0.1686
I am getting a wrong answer for b. Please help. My thoughts:
(a) is dead easy. Drawing the sample spaces, there are four scenarios. Let H denote a hit and M denote a miss. The four scenarios are:
Sum of the probabilities of these four cases turns out to be exactly 0.6976 (lots of small decimal multiplications!), so all is good. I've got the correct answer.
For the second part, my approach is:
There are two cases in our favor:
Case 1: MMH (3 shots exactly)
Case 2: MMMH (4 shots)
Adding the probabilities for these two cases, I get 0.1176 as the answer. But this is wrong according the to solution index. What am I doing wrong?
Also, can someone be kind enough to show me how to model the second part using Bayes theorem of conditional probability? In the form $P(>=3|S)$, i.e. the probability of at least 3 shots being used, knowing already that the plane was shot.
In particular, I understand that $P(>=3|S) = P(>3 \cap S) / P(S)$. We have calculated $P(S)$ in part a. I don't know how to calculate the intersection in the numerator.