Probability Part 1

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⇒‌

x

z

=6
Now, ‌

P(B1)

P(B3)

=‌

x

z

=6
∴ Required answer is 6.