1
$\begingroup$

Of the three independent events $E_1,E_2,E_3$,the probability that only $E_1$ occurs is $\alpha$,only $E_2$ is $\beta$ and only $E_3$ is $\gamma$. Let the probability $p$ that none of the events occurs satisfy the equations $(\alpha-2\beta)p=\alpha\beta$ and $(\beta-3\gamma)p=2\beta\gamma$. Probability of occurence of $E_1$ is$X$ and probability of occurence of $E_3$ is $Y$. Find $\frac{X}{Y}$. What can i do to proceed?I'm totally strucked!

$\endgroup$

1 Answer 1

1
$\begingroup$

Let $\delta, \epsilon, \zeta$ be the probabilities that $E_1$ and $E_2$, $E_1$ and $E_3$, and $E_2$ and $E_3$ occur, respectively. Let $\eta$ be the probability that all three occur.

Then

$$p = 1 - \alpha - \beta - \gamma + \delta + \epsilon + \zeta - \eta.$$

Also, $X = \alpha + \delta + \epsilon - \eta,$ and $Y = \beta + \delta + \zeta - \eta.$

Can you take it from there?

$\endgroup$
1
  • $\begingroup$ Thanks a lot!Yup i can take it ahead. $\endgroup$ Commented Jan 31, 2014 at 17:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .