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Probability Level 3

In a test, an examine either guesses or copies or knows the answer to a multiple choice question with four choices. The Probability that he makes a guess is 1 3 \dfrac{1}{3} and the probability that he copies the answer is 1 6 . \dfrac{1}{6}. The probability that his answer is correct, given that he copied it, is 1 8 \dfrac{1}{8} . Find the probability that he knew the answer to the question, given that he correctly answered it.

  • If the probability can be written as a b \dfrac{a}{b} , where a a and b b are coprime positive integers,
  • Find a × b a\times b .


The answer is 696.

Samara Simha Reddy by Samara Simha Reddy , 22, India

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1 solution

E 1 = Examine guesses the answer , E 2 = Examine copies the answer , E 3 = Examine knows the answer and A = Examine answers correctly Clearly, P ( E 1 ) = 1 3 , P ( E 2 ) = 1 6 . ∵ E 1 , E 2 , E 3 are mutually exclusive and exhaustive events. ∴ P ( E 1 ) + P ( E 2 ) + P ( E 3 ) = 1 ⟹ P ( E 3 ) = 1 2 P ( A / E 1 ) = 1 4 , P ( A / E 2 ) = 1 8 , P ( A / E 3 ) = 1. ∴ Required Probability = P ( E 3 / A ) = P ( E 3 ) . P ( A / E 3 ) P ( E 1 ) . P ( A / E 1 ) + P ( E 2 ) . P ( A / E 2 ) + P ( E 3 ) . P ( A / E 3 ) = 1 2 × 1 1 3 × 1 4 + 1 6 × 1 8 + 1 2 × 1 . = 24 29 ⟹ a b = 24 29 a × b = 24 × 29 = 696 . \large \displaystyle E_1 = \text{ Examine guesses the answer },\\ \large \displaystyle E_2 = \text{ Examine copies the answer },\\ \large \displaystyle E_3 = \text{ Examine knows the answer and } A = \text{ Examine answers correctly }\\ \large \displaystyle \text{Clearly, } P(E_1) = \frac{1}{3}, P(E_2) = \frac{1}{6}.\\ \large \displaystyle \because E_1, E_2, E_3 \text{are mutually exclusive and exhaustive events. }\\ \large \displaystyle \therefore P(E_1) + P(E_2) + P(E_3) = 1 \implies P(E_3) = \frac{1}{2}\\ \large \displaystyle P(A/E_1) = \frac{1}{4}, P(A/E_2) = \frac{1}{8}, P(A/E_3) = 1.\\ \large \displaystyle \therefore \text{Required Probability } = P(E_3/A)\\ \large \displaystyle = \frac{P(E_3) . P(A/E_3)}{P(E_1) . P(A/E_1) + P(E_2) . P(A/E_2) + P(E_3) . P(A/E_3)}\\ \large \displaystyle = \frac{\frac{1}{2} \times 1}{\frac{1}{3} \times \frac{1}{4} + \frac{1}{6} \times \frac{1}{8} + \frac{1}{2} \times 1}. = \frac{24}{29}\\ \large \displaystyle \implies \frac{a}{b} = \frac{24}{29}\\ \large \displaystyle a \times b = 24 \times 29 = \color{#3D99F6}{\boxed{696}}.

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