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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:

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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4 Answers 4

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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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HINT:

a conditional probability is dependent on a condition. What is the condition here? 3 or four shots to hit the plane implies that the plane gets hit.

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  • $\begingroup$ It's told upfront that the plane is hit, not that we decipher it from the fact that it takes three or four shots to hit the plane (I mean, of course what you've said is absolutely correct, but I prefer to be forward conclusive with the thought process, if you know what I mean). So speaking in conditional terms, GIVEN that the plane was hit, what's the probability that it took 3 or 4 shots? So I was basically missing that my cases were limited to hit cases (0.6976), and the two favoring cases I had already calculated (0.1176). That's it. $\endgroup$
    – IntelligentMoron
    Jul 31, 2011 at 19:12
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There is an easier method of computing part a. The probability that the plane is not hit is .6*.7*.8*.9=.3024, 1-.3024=.6976. For part b you obtained .1176. However this has to be divided by the probability that the plane actually gets hit (.6976),=> .1176/.6976=.1686

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This is a very Basic Probability Question. These methods are Verified.

Please read carefully to understand.

(a) What is the probability that the gun is able to hit the plane?

This means, P(Hitting the Target At least once) = ?

$Method$ $1 :$

GIVEN in the problem: The probability of hitting the plane at $p1 = 1st, p2 = 2nd, p3=3rd$ and $p4 = 4th$ shots are $0.4, 0.3, 0.2$ and $0.1$ respectively.

Then,

P(Gun HITS the plane ) = P(The plane is HIT AT LEAST Once) = $1 - $ P(The plane is HIT IN NONE of the shots) $ = $1 -$ [(1- p1)(1- p2)(1- p3)(1- p4)]$ = $1 - [(1 - 0.4)(1 - 0.3)(1 - 0.2)(1 - 0.1)] = 1 - [0.6* 0.7 * 0.8 * 0.9 ] = [1 - 0.3024] = 0.6976 $

$Method$ $2:$

If The information about the Probability of hitting the plane at different shots is not given, then also you can use this method.

Formulae to be Used : Sum of Nth terms of G.P : $S_n = a\frac{(1 – r^{n})}{(1 – r )}$

Given : aircraft gun fires at a moving enemy plane with 4 successive shots.

Let's say, he fires at the target K times. Then, He can Kill his enemy by his 1st Hit, 2nd Hit, 3rd Hit, and so on.

$P(e)$ => Man Successfully HITS the target

$P(\bar{e})$ => Man Failing to Hit the target

Succcess OR Fail Succcess OR Fail Fail Succcess OR... and so on can be written as : $P(e) + P(\bar{e})P(e) + P(\bar{e})P(\bar{e})P(e) + .... $ and so on

Therefore, P(Hitting the Target At least once) = $ P(e) + P(\bar{e})P(e) + P(\bar{e})P(\bar{e})P(e) + ..... [P(\bar{e})....(K-1) times ]P(e) = P(e) [ 1 + P(\bar{e}) + P(\bar{e})^{2} + P(\bar{e})^{3} + ... + P(\bar{e})^{K - 1 } ] = [\frac{1 - P(\bar{e})^{K}}{1 - P(\bar{e})}]P(e) = [\frac{1 - P(\bar{e})^{K}}{P(e)}]P(e) = [1 - P(\bar{e})^{K} )] = [1 - (1 - 1/4)^{4}] = 1 - (3/4)^{4} = 1 - 0.31640625=0.68359375 $

Now, your $2nd$ $Question$...

Question 2 : What is the probability of his Hitting the Target at least Thrice ?

Solution of Question 2:

$n = 4$

$ Success = p = 1/4$

$Failure = F = q = 1- 1/4 = 3/4$

We will use the formulae : $\binom{n}{r}p^{r}q^{n-r}$ and subtract 1 from it.

$ P(F≥3) = 1 −[P(F= 0) + P(F= 1)+ P(F= 2)] = 1 - [\binom{4}{0}.(\frac{1}{4})^{0}.(\frac{3}{4})^{4-0} + \binom{4}{1}.(\frac{1}{4})^{1}.(\frac{3}{4})^{4-1} ++ \binom{4}{2}.(\frac{1}{4})^{2}.(\frac{3}{4})^{4-2} ]=1-[\binom{4}{0}.(\frac{1}{4})^{0}.(\frac{3}{4})^{4} + \binom{4}{1}.(\frac{1}{4})^{1}.(\frac{3}{4})^{3} + \binom{4}{2}.(\frac{1}{4})^{2}.(\frac{3}{4})^{2} ] $

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