# Elementary Probability Question: Conditional case, with AT LEAST clause

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?

(b) What is the conditional probability that at least three shots are required to be shot?

(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:

1: H
2. MH
3. MMH
4. MMMH

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.

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The question asks for the conditional probability that at least three shots were required to be shot. This means what is the chance that either three or four shots were fired and hit the plane, given that the plane was hit. This is just equal to the probability that it takes three or four shots, namely the 0.1176 that you calculated, divided by the probability that the gun hits the plane, 0.6976. When you divide the two, you'll notice that you get 0.1686.

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