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

---

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

---

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

another way of solving is the question says that convict not getting injured is .5 and p(x)=3p(y)

so.. (1-p(x))(1-p(y))=.5

(1-3p(y))(1-p(y))=.5

solving it we will get the same equation.. 6t²-8t+1=0..

so ans is .14

---

Yeah, thats also correct :D
we get the same equation.
I think my method was longer :P
I will add your solution as alternate one.

---

Another one:

P(Z)=P(X∪Y) i.e. when either X or Y hits the target

P(X)=3*P(Y)

Let P(Y)=y then P(X)=3y

P(X∪Y)=P(X) + P(Y) + P(X∩Y)

P(X∩Y)=P(X)*P(Y) as X and Y are independent events.

=>0.5=x+3y+y*3y => 3y²-4y+0.5=0

Solving the quadratic equation we will get y=0.1396 ≌0.14

 

---

@MiNiPanda

Can u please give a Tree for it ..

---

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

---

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