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Bunuel
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Three persons P, Q and R independently try to hit a target. If the pro 01 Feb 2022, 04:00
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A
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65% (hard)

Question Stats:

59% (01:59) correct 41% (01:54) wrong based on 115 sessions

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Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are 3/4, 1/2 and 5/8 respectively, then the probability that the target is hit by P or Q but not by R is

A) 21/64
B) 9/64
C) 15/64
D) 39/64
E) 27/64
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A
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Three persons P, Q and R independently try to hit a target. If the pro 01 Feb 2022, 08:33
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Bunuel
Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are 3/4, 1/2 and 5/8 respectively, then the probability that the target is hit by P or Q but not by R is

A) 21/64
B) 9/64
C) 15/64
D) 39/64
E) 27/64
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If P(person P hits) = 3/4, then P(person P misses) = 1 - 3/4 = 1/4
If P(person Q hits) = 1/2, then P(person P misses) = 1 - 1/2 = 1/2
If P(person R hits) = 5/8, then P(person P misses) = 1 - 5/8 = 3/8

We want to find P(target is hit by P or Q but not by R)
There are three different scenarios that satisfy this:
1) P hits, Q misses, and R misses
2) P misses, Q hits, and R misses
3) P hits, Q hits, and R misses

So, P(target is hit by P or Q but not by R) = P(P hits AND Q misses AND R misses OR P misses AND Q hits AND R misses OR P hits AND Q hits AND R misses)
= P(P hits AND Q misses AND R misses) + P(P misses AND Q hits AND R misses) + P(P hits AND Q hits AND R misses)
= [P(P hits) x P(Q misses) x P(R misses)] + [P(P misses) x P(Q hits) x P(R misses)] + [P(P hits) x P(Q hits) x P(R misses)]
= [3/4 x 1/2 x 3/8] + [1/4 x 1/2 x 3/8] + [3/4 x 1/2 x 3/8)]
= 9/64 + 3/64 + 9/64
= 21/64

Answer: A
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Three persons P, Q and R independently try to hit a target. If the pro 01 Feb 2022, 14:31
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Bunuel
Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are 3/4, 1/2 and 5/8 respectively, then the probability that the target is hit by P or Q but not by R is

A) 21/64
B) 9/64
C) 15/64
D) 39/64
E) 27/64
Show more

Probability of P OR Q is

Pr(P)+Pr(Q)-Pr(P and Q), equaling

3/4 + 1/2 - 3/4*1/2 = 5/4 - 3/8=

7/8

This is coupled with R not hitting, with a probability of

1 - 5/8 = 3/8

So the total probability is

7/8 * 3/8 = 21/64

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Three persons P, Q and R independently try to hit a target. If the pro 11 May 2024, 11:22
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BrentGMATPrepNow

Bunuel
Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are 3/4, 1/2 and 5/8 respectively, then the probability that the target is hit by P or Q but not by R is

A) 21/64
B) 9/64
C) 15/64
D) 39/64
E) 27/64
If P(person P hits) = 3/4, then P(person P misses) = 1 - 3/4 = 1/4
If P(person Q hits) = 1/2, then P(person P misses) = 1 - 1/2 = 1/2
If P(person R hits) = 5/8, then P(person P misses) = 1 - 5/8 = 3/8

We want to find P(target is hit by P or Q but not by R)
There are three different scenarios that satisfy this:
1) P hits, Q misses, and R misses
2) P misses, Q hits, and R misses
3) P hits, Q hits, and R misses

So, P(target is hit by P or Q but not by R) = P(P hits AND Q misses AND R misses OR P misses AND Q hits AND R misses OR P hits AND Q hits AND R misses)
= P(P hits AND Q misses AND R misses) + P(P misses AND Q hits AND R misses) + P(P hits AND Q hits AND R misses)
= [P(P hits) x P(Q misses) [b]x P(R misses)] + [P(P misses) x P(Q hits) [b]x P(R misses)] + [P(P hits) x P(Q hits) [b]x P(R misses)]
= [3/4 x 1/2 x 3/8] + [1/4 x 1/2 x 3/8] + [3/4 x 1/2 x 3/8)]
= 9/64 + 3/64 + 9/64
= 21/64

Answer: A
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­Its mentioned in question "P or R" right then why have we considered scenario of both P and Q hitting 
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