Probability of solving a priblem of a,b,c are 1/3, 2/7, and 3/8 respectively. if all the three try to solve the problem simultaneously, find the probability that exactly one of them can solve it
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We have P(A) = 1/3 => P(not A) = 2/3,
P(B) = 2/7 => P(not B) = 5/7,
P(C) =3/8 => P(not C) = 5/8.
Probability that only one will solve the problem
= P(A)and P(not B) and P(not C) + P(not A) and P(B) and P(not C)+ P(not A) and P(not B) and P(C)
= 1/3×5/7×5/8 + 2/3×2/7×5/8 + 2/3×5/7×3/8
= (25/168) + (20/168) + (30/168)
= 75/168 = 25/56. [Ans.]
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P(B) = 2/7 => P(not B) = 5/7,
P(C) =3/8 => P(not C) = 5/8.
Probability that only one will solve the problem
= P(A)and P(not B) and P(not C) + P(not A) and P(B) and P(not C)+ P(not A) and P(not B) and P(C)
= 1/3×5/7×5/8 + 2/3×2/7×5/8 + 2/3×5/7×3/8
= (25/168) + (20/168) + (30/168)
= 75/168 = 25/56. [Ans.]
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