Of the three independent event `E_(1),E_(2)` and `E_(3)`, the probability that only `E_(1)` occurs is `alpha`, only `E_(2)` occurs is `beta` and only

Question

Of the three independent event E1,E2 and E3 , the probability that only E1 occurs is α , only E2 occurs is β and only E3 occurs is γ . If the probavvility p that none of events E1,E2 or E3 occurs satisfy the equations (α-2β)p=αβ and (β-3γ)p=2βγ . All the given probabilities are assumed to lie in the interval (0, 1). Then, probability of occurrence of E1 probability of occurrence of E3 is equal to A. 3 B. 2 C. 6 D. 4 Select the correct answer from above options

Answer 1

Correct Answer - C Let P(E1)=x,P(E2)=y and P(E3)=z It is given that : P(Only E1 occurs)=α ⇒P(E1∩ ¯ E2 ∩ ¯ E3 )=α ⇒P(E1)P( ¯ E2 )P(E3)=α ⇒x(1-y)(1-z)=α P(Only E2 occurs)=β ⇒P( ¯ E1 ∩E2∩ ¯ E3 )=β ⇒P( ¯ E1 )P(E2)P( ¯ E3 )=β ⇒(1-x)y(1-z)=β P(only E3 occurs)=γ ⇒P( ¯ E1 ∩ ¯ E2 ∩E3)=γ ⇒P( ¯ E1 )P( ¯ E2 )P(E3)=γ ⇒P( ¯ E1 )P( ¯ E2 )P(E3)=γ ⇒(1-x)(1-y)z=γ and, P(None of the events E1,E2 and E3 occurs)=p ⇒P( ¯ E1 ∩ ¯ E2 ∩ ¯ E3 )=p ⇒P( ¯ E1 )P( ¯ E2 )P( ¯ E3 )=p ⇒(1-x)(1-y)(1-z)=p Substituting these values in (α-2β)p=αβ and (β-3γ)p=2βγ , we obtain x=2y and y=3z. These two equations give x=6z. Hence, P(E1) P(E3) = x z =6