Given, B1,B2 and B3 are three independent events. Let x,y,z be the probability of B1,B2,B3′ respectively. P(only B1 occur )=α P(B1)⋅P(B2)⋅P(B3)=α ⇒x⋅(1−y)⋅(1−z)=α ⇒ P(only B2 occur )=β P(B1)⋅P(B2)⋅P(B3)=β ⇒(1−x)⋅y⋅(1−z)=β P(none occur) =P P(B1)⋅P(B2)⋅P(B3)=P ⇒(1−x)⋅(1−y)⋅(1−z)=P Now, we have given relations (α−2β)P=αβ ⇒[x(1−y)(1−z)−2y(1−x)(1−z)] (1−x)(1−y)(1−z) =x⋅(1−y)(1−z)⋅y(1−x)(1−z) [ putting the value of α,β,P] ⇒(1−z)[x(1−y)−2y(1−x)]=x⋅y⋅(1−z) ⇒x−xy−2y+2xy=xy ⇒x=2y . . . (i) Similarly, on solving the second relation, (β−3γ)P=2βγ by putting β,γ and P, We get, y=3z . . . (ii) From Eqs. (i) and (ii), we get x=2×3z⇒x=6z ⇒