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( only B3 occur )=γ ⇒P(B1)⋅P(B2)⋅P(B3)=γ ⇒(1−x)⋅(1−y)⋅z=γ 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) ⇒(1−z)[x(1−y)−2y(1−x)]=x⋅y⋅(1−z) ⇒x−xy−2y+2xy=xy ⇒x=2y Similarly, on solving the second relation, (β−3γ)P=2βγ by putting β,γ and P, we get y=3zx=2×3z From Eqs. (i) and (ii), we get ⇒x=6z⇒
