Question

Let there by three independent events $E_1,E_2$ and $E_3$. The probability that only $E_1$ occurs is $\alpha$ , only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$. Let ${}^{\prime }{p}^{\prime }$ denote the probability of none of events occurs that satisfies the equations $(\alpha -2\beta )p=\alpha \beta$ and $(\beta -3\gamma )p=2\beta \gamma$. All the given probabilities are assumed to lie in the interval $(0,1)$
Then, $\frac{\text{probability of occurrence of}{E}_1}{\text{probability of occurrence of}{E}_3}$ is equal to

• A. 6.0
• B. 6.00
• C. 6
• D. 06

Answer 1

Answer: (A), (B), (C), (D)

Let $x,y$ and $z$ be probability of $E_1,E_2$ and $E_3$ respectively, then
$\Rightarrow x(1–y)(1–z)=\alpha \Rightarrow y(1–x)(1–z)=\beta \Rightarrow z(1–x)(1–y)=\gamma \Rightarrow (1–x)(1–y)(1–z)=p$
Given $(\alpha -2\beta )p=\alpha \beta$
$\Rightarrow \alpha -2\beta =\frac{\alpha \beta }{p}\Rightarrow (1-z)[x-xy-2y+2xy]=xy(1-z)\Rightarrow x=2y$
Given $(\beta -3\gamma )p=2\beta \gamma$
$\Rightarrow \beta -3\gamma =\frac{2\beta \gamma }{p}\Rightarrow (1-x)[y-yz-3z+3yz]=2yz(1-x)\Rightarrow y=3z\therefore x=2y=6z$

Hence,
$\frac{\text{probability of occurrence of}{E}_1}{\text{probability of occurrence of}{E}_3}=\frac{x}{z}=6$