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# Three persons P, Q and R independently try to hit a target.

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Manager
Joined: 27 May 2010
Posts: 200
Three persons P, Q and R independently try to hit a target.  [#permalink]

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25 Jun 2019, 02:11
00:00

Difficulty:

55% (hard)

Question Stats:

60% (01:52) correct 40% (01:37) wrong based on 35 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 & 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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Intern
Joined: 24 May 2019
Posts: 1
Location: India
Three persons P, Q and R independently try to hit a target.  [#permalink]

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25 Jun 2019, 16:17
Probability of p or q but not r =(3/4)(1/2)(3/8)+(1/4)(1/2)(3/8)+(3/4)(1/2)(3/8)= 21/64

A

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Senior Manager
Joined: 15 Feb 2018
Posts: 299
Re: Three persons P, Q and R independently try to hit a target.  [#permalink]

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07 Aug 2019, 16:20
1
I used the formula $$P(AorB)=P(A)+P(B)-P(AB)$$
The above formula needed to be modified to also allow for R missing

Use 1-P for R missing, $$1-\frac{5}{8}=\frac{3}{8}$$

$$P(PorQnotR)=P(PnotR)+P(QnotR)-P(PQnotR)$$
$$P(PorQnotR)=\frac{3}{4}*\frac{3}{8}+\frac{1}{2}*\frac{3}{8}-\frac{3}{4}*\frac{1}{2}*\frac{3}{8}$$
$$P(PorQnotR)=\frac{9}{32}+\frac{3}{16}-\frac{9}{64}$$
$$P(PorQnotR)=\frac{18}{64}+\frac{12}{64}-\frac{9}{64}$$
$$P(PorQnotR)=\frac{21}{64}$$

A
Re: Three persons P, Q and R independently try to hit a target.   [#permalink] 07 Aug 2019, 16:20
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