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Question
Read the following passage and answer the questions given below.
 There are two antiaircraft guns, named as A and B. The probabilities that the shell fired from them hits an airplane are 0.3 and 0.2 respectively. Both of them fired one shell at an airplane at the same time. |
- What is the probability that the shell fired from exactly one of them hit the plane?
- If it is known that the shell fired from exactly one of them hit the plane, then what is the probability that it was fired from B?
Solution
i. Let P be the event that the shell fired from A hits the plane and Q be the event that the shell fired from B hits the plane.
The following four hypotheses are possible before the trial, with the guns operating independently:
`E_1 = PQ, E_2 = barP barQ, E_3 = barP Q, E_4 = P barQ`
Let E = The shell fired from exactly one of them hits the plane.
P(E1) = 0.3 × 0.2 = 0.06, P(E2) = 0.7 × 0.8 = 0.56, P(E3) = 0.7 × 0.2 = 0.14, P(E4) = 0.3 × 0.8 = 0.24
`P(E/E_1)` = 0, `P(E/E_2)` = 0, `P(E/E_3)` = 1, `P(E/E_4)` = 1
P(E) = `P(E_1). P(E/E_1) + P(E_2). P(E/E_2) + P(E_3). P(E/E_3) + P(E_4). P(E/E_4)`
= 0.14 + 0.24
= 0.38
ii. By Bayes’ Theorem, `P(E_3/E) = (P(E_3).P(E/E_3))/(P(E_1).P(E/E_1) + P(E_2).P(E/E_2) + P(E_3).P(E/E_3) + P(E_4).P(E/E_4))`
= `0.14/0.38`
= `7/19`
NOTE: The four hypotheses form the partition of the sample space and it can be seen that the sum of their probabilities is 1. The hypotheses E1 and E2 are actually eliminated as `P(E/E_1) = P(E/E_2)` = 0
Alternative way of writing the solution:
i. P(Shell fired from exactly one of them hits the plane)
= P[(Shell from A hits the plane and Shell from B does not hit the plane) or (Shell from A does not hit the plane and Shell from B hits the plane)]
= 0.3 × 0.8 + 0.7 × 0.2
= 0.38
ii. P(Shell fired from B hit the plane/Exactly one of them hit the plane)
= `(("P(Shell fired from B hit the plane" ∩ "Exactly one of them hit the plane"))/("P(Exactly one of them hit the plane")`
= `(("P(Shell from only B hit plane"))/("P(Exactly one of them hit the plane")`
= `0.14/0.38`
= `7/19`
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