Answer:
Option D
Explanation:
Concept involved
For the events to be independent.
P(E1∩E2∩E3)=P(E1)P(E2)P(E3)
P(E1∩¯E2∩¯E3)=P
(on;y E1 occurs)
= P(E1).(1−P(E2))(1−P(E3))
let x,y,z are probability of E1, E2 and E3 respectively
∴ α = x(1-y)(1-z) ......(i)
β = (1-x).y(1-z) ......(ii)
γ = (1-x)(1-y) z ..........(iii)
⇒ p = (1-x) (1-y) (1-z) ..........(iv)
Given (α−2β)p=αβ
and (β−3γ)p=2βγ .......(v)
From above equations
x=2y and y=3 z
∴ x=6z = xz=6