P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, tw

Question

P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit?

A. 1/6
B. 1/3
C. 1/2
D. 2/3
E. 5/6

eswarchethu135 · Accepted Answer

P (hitting the target) =  \frac{4}{5} 
P (not hitting the target) =  \frac{1}{5}

Q (hitting the target) =  \frac{3}{4} 
Q (not hitting the target) =  \frac{1}{4}

R (hitting the target) = \frac{2}{3} 
R (not hitting the target) = \frac{1}{3}

Required probability is  PQ\overline{R}  +  P\overline{Q}R  +  \overline{P}QR  +  PQR

=  \frac{1}{5} + \frac{1}{10} + \frac{2}{15} + \frac{2}{5}

=  \frac{5}{6}

OPTION:  E

Archit3110 · Answer

P of P hit= 4/5 P miss 1/5
P of Q hit = 3/4 , miss 1/4
P of R hit 2/3 , miss 1/3

P of atleast 2 hits 
(4/5*3/4*1/3)+(4/5*1/4*2/3)+(1/5*3/4*2/3)+(4/5+3/4+2/3)

Solve we get option E ;5/6

IanStewart · Answer

This is not how probability questions are worded -- if we say something like "a car can drive 80 miles in one hour", we're describing a maximum, not a probability.

Assuming the question is expressing probabilities, each individual probability is so high that only answers D and E are even remotely plausible. You can avoid working out all the cases here just by asking "what is the probability that P and one other shooter hits the target?" That probability alone is higher than 2/3, so E must be right.

tanvi0915 · Answer

P hit = 4/5 P miss = 1/5
Q hit = 3/4 P miss = 1/4
R hit = 2/3 R miss = 1/3
List down all the cases possible for atleast 2 shots
P hit Q hit R miss
P hit Q miss R hit
P miss Q hit R hit
P hit Q hit R hit

Now Calculate the prob
(4/5*3/4*1/3) + (1/5*3/4*2/3) + (4/5*1/4*2/3) + (4/5*3/4*2/3) = (12+6+8+24)/60 = 5/6
Answer is E

P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, tw

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P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, tw

Prep Toolkit

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