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13 Nov 2018, 10:48
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C
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Difficulty:
25%
(medium)
Question Stats:
81% (02:06) correct
19%
(01:40)
wrong
based on 167
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In a shooting competition, probability of A hitting target is 2/5, by B is 2/3 and C is 3/5. If all of them fire independently, what is the probability that only one will hit target?
(A) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4
(A) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4
13 Nov 2018, 11:07
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P(x) = probability of hitting target, P(x") = probability of not hitting target = 1 - P(x), using this:
P(A) = 2/5, so, P(A") = 3/5
P(B) = 2/3, so, P(B") = 1/3
P(C) = 3/5, so, P(C") = 2/5
OR = plus, and = *
We need: P(A hits & B doesnt & C doesnt) OR P(A doesnt & B hits & C doesnt) OR P(A doesnt & B doesnt & C hits)
which is: P(A)(B")(C") + P(A")P(B)P(C") + P(A")P(B")P(C)
= \((2/5)*(1/3)*(2/5) + (3/5)*(2/3)*(2/5) + (3/5)*(1/3)*(3/5)\)
= \(\frac{(4+12+9)}{75}\)
= 25/75 = 1/3
Option C
P(A) = 2/5, so, P(A") = 3/5
P(B) = 2/3, so, P(B") = 1/3
P(C) = 3/5, so, P(C") = 2/5
OR = plus, and = *
We need: P(A hits & B doesnt & C doesnt) OR P(A doesnt & B hits & C doesnt) OR P(A doesnt & B doesnt & C hits)
which is: P(A)(B")(C") + P(A")P(B)P(C") + P(A")P(B")P(C)
= \((2/5)*(1/3)*(2/5) + (3/5)*(2/3)*(2/5) + (3/5)*(1/3)*(3/5)\)
= \(\frac{(4+12+9)}{75}\)
= 25/75 = 1/3
Option C
14 Nov 2018, 02:10
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Bunuel
P (A)= 2/5 , Pn(A)= 3/5
P(B)= 2/3, Pn(B)=1/3
P(C)= 3/5, Pn(C)=2/5
only one will hit Probability=
{2/5*1/3*2/5}+{3/5*2/3*2/5}+{3/5*1/3*3/5}
= 25/75
=1/3 option C
