Let there be 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 $p'$ denote the probability of none of events occurs that satisfies the equations $(\alpha-2\beta)p=\alpha\beta$ and…

Question

Let there be 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 $p'$ 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, the $\frac{	ext{Probability of occurrence of } E_1}{	ext{Probability of occurrence of } E_3}$ is equal to ________.

Aicharya · Accepted Answer

Let P E 1 = P 1 ; P E 2 = P 2 ; P E 3 = P 3 P E 1 &#8745; E &#175; 2 &#8745; E &#175; 3 = &#945; = P 1 1 - P 2 1 - P 3 &#160; &#8230; 1 P E &#175; 1 &#8745; E 2 &#8745; E &#175; 3 = &#946; = 1 - P 1 P 2 1 - P 3 &#160; &#8230; 2 P E &#175; 1 &#8745; E &#175; 2 &#8745; E 3 = &#947; = 1 - P 1 1 - P 2 P 3 &#160; &#8230; 3 P E &#175; 1 &#8745; E &#175; 2 &#8745; E &#175; 3 = p = 1 - P 1 1 - P 2 1 - P 3 &#160; &#8230; 4 Given that, ( &#945; - 2 &#946; ) p = &#945; &#946; &#8658; P 1 1 - P 2 1 - P 3 - 2 1 - P 1 P 2 1 - P 3 p = P 1 P 2 1 - P 1 1 - P 2 1 - P 3 2 &#8658; P 1 1 - P 2 - 2 1 - P 1 P 2 = P 1 P 2 &#8658; P 1 - P 1 P 2 - 2 P 2 + 2 P 1 P 2 = P 1 P 2 &#8658; P 1 = 2 P 2 &#160; &#8230; 1 and similarly, ( &#946; - 3 &#947; ) P = 2 &#946; &#947; P 2 = 3 P 3 &#160; &#8230; 2 So, P 1 = 6 P 3 &#8658; P 1 P 3 = 6

Let there be three independent events $E_1$, $E_2$ and $E_3$. The probability that only $E_1$ occurs is…

Solution