Question

Let EandF be tow independent events. The probability that exactly one of them occurs is 11/25 and the probability if none of them occurring is 2/25. If P(T) deontes the probability of occurrence of the event T, then P(E)= 4 5 ,P(F)= 3 5 P(E)= 1 5 ,P(F)= 2 5 P(E)= 2 5 ,P(F)= 1 5 P(E)= 3 5 ,P(F)= 4 5 A. P(E)= 4 5 ,P(F)= 3 5 B. P(E)= 1 5 ,P(F)= 2 5 C. P(E)= 2 5 ,P(F)= 1 5 D. P(E)= 6 5 ,P(F)= 1 5 Select the correct answer from above options

Answer 1

Correct Answer - A We have, P(E)+P(F)-2P(E∩F)= 11 25 andP( ˉ E ∩ ˉ F )= 2 25 ⇒P(E)+P(F)-2P(E)P(F)= 11 25 and P( ˉ E )P( ˉ F )= 2 25 ⇒x+y-2xy= 11 25 and 1-x+xy= 2 25 , where P(E ) =x and P(F)=y ⇒x+y+2-2x-2y= 11 25 +2× 2 25 [ On eliminating xy] ⇒x+y= 7 5 ⇒y= 7 5 -x Substituting y= 7 5 -x in 1-x-y+xy= 2 25 , we get 1- 7 5 +x( 7 5 -x)= 2 25 ⇒25x2-35x+12=0⇒x= 3 5 , 4 5 When x= 3 5 ,y= 4 5 and y= 3 5 for x= 4 5 Hence, P(E)= 3 5 ,P(F)= 4 5 or P(E)= 4 5 ,P(F)= 3 5