GATE2012 AE: GA-6

Question

Two policemen, $A$ and $B$, fire once each at the same time at an escaping convict. The probability that $A$ hits the convict is three times the probability that $B$ hits the convict. If the probability of the convict not getting injured is $0.5$, the probability that $B$ hits the convict is

• A. $0.14$
• B. $0.22$
• C. $0.33$
• D. $0.40$

Comments

A

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Please Anyone Provide some different approach.  .

Answer 1

$X-$ $A$ hits the convict
$Y -$ $B$ hits the convict
Given, $P(X) = 3 \times P(Y)$

$Z -$ Convict is injured
$Z' -$ Convict is not injured

Given, $P(Z') = 0.5$
$\implies P(Z) = 1 - P(Z') = 1 - 0.5 = 0.5$

Now,
$P(Z) = P(X) \times P(Y') + P(X') \times P(Y) + P(X) \times P(Y)$
Let $P(Y) = t$
$P(X) = 3t$
$P(Y') = 1-t$
$P(X') = 1-3t$

Substituting in above equation,
$0.5 = (3t \times (1-t)) +  ((1-3t) \times t) + (t \times 3t)$
$\implies 3t - 3t^2 + t - 3t^2 + 3t^2 = 0.5$
$\implies 3t^2 - 4t + 0.5 = 0$
$\implies 6t^2 - 8t + 1 = 0$

Solving, we get $t=1.193$ (eliminated as probability cannot be greater than $1$) OR $t=0.1396$

Therefore. $P(Y) = t = 0.1396$
Answer (A) 0.14

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Alternative Method: by Joker 
$P(Z')=0.5$
Now,
$P(Z') = (P(X') \times P(Y'))$
$\implies 0.5 = (1-3t) \times (1-t)$

Solving this gives the same equation as above.
$6t^2 - 8t + 1 = 0$
and the same answer.

Comments

@MiNiPanda

Can u please give a Tree for it ..

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It is not based on BAYES Theorem... It is just independent events

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@MiNiPanda P(X∪Y)=P(X) + P(Y) - P(X∩Y), you have added the intersection part.