B_i (i=1,2,3) are independent events; only B_1 occurs with probability , only B_2 with , o…

Question

$B_i$ ($i=1,2,3$) are independent events; only $B_1$ occurs with probability $\alpha$, only $B_2$ with $\beta$, only $B_3$ with $\gamma$. Let $P$ = probability none occurs, satisfying $(\alpha-2\beta)P=\alpha\beta$ and $(\beta-3\gamma)P=2\beta\gamma$. Find $\dfrac{P(B_1)}{P(B_3)}$.

Answer: 6

Accepted Answer

Let $P(B_1)=x,P(B_2)=y,P(B_3)=z$. Then $\alpha=x(1-y)(1-z)$, $\beta=(1-x)y(1-z)$, $P=(1-x)(1-y)(1-z)$.

From $(\alpha-2\beta)P=\alpha\beta$, simplifying gives $x=2y$. From $(\beta-3\gamma)P=2\beta\gamma$, similarly $y=3z$.

So $x=6z \Rightarrow \dfrac{P(B_1)}{P(B_3)}=6$.

Solutions · 1