hi..please help me to solve...The discrete random variable x can only take the values of 1,2,3 and 4,with the probabilities p(x=1)=p(x=2)and p(x=3)=p(x=4)=2p(x=1).If A={1,3},find p(A).

Printable View

- September 21st 2012, 09:51 PMsharmalaplease help me to solve this probability question.....
hi..please help me to solve...The discrete random variable x can only take the values of 1,2,3 and 4,with the probabilities p(x=1)=p(x=2)and p(x=3)=p(x=4)=2p(x=1).If A={1,3},find p(A).

- September 21st 2012, 10:14 PMchiroRe: please help me to solve this probability question.....
Hey sharmala.

First you need to find the probabilities. We know that the sum of all probabilities has to equal 1 and you have four probabilities where p1 + p2 + p3 + p4 = 1, but you have a relationship to link all these together.

Also since A contains two events the probability of A is P(A) = P(X=1 OR X=3).

I'll wait for you to make an attempt to solve the problem and give some hints but I need your feedback on what your thinking and what you have tried. - September 21st 2012, 10:34 PMsharmalaRe: please help me to solve this probability question.....
Hey sharmala.

First you need to find the probabilities. We know that the sum of all probabilities has to equal 1 and you have four probabilities where p1 + p2 + p3 + p4 = 1, but you have a relationship to link all these together.

Also since A contains two events the probability of A is P(A) = P(X=1 OR X=3).

I'll wait for you to make an attempt to solve the problem and give some hints but I need your feedback on what your thinking and what you have tried.

Thanks

Reply Reply With Quote

hai...im still cant catch up...please give me some hint.. - September 21st 2012, 10:36 PMchiroRe: please help me to solve this probability question.....
First Hint: p1 = P(X=1), p2 = P(X=2), p3 = P(X=3) and p4 = P(X=4). Now you have an expression in terms of these and you know that p1 + p2 + p3 + p4 = 1, but you can write p2 in terms of p1, p3 in terms of p1 and p4 in terms of p1. So using this can you find p1 and hence find p2, p3, and p4?

- September 21st 2012, 11:21 PMsharmalaRe: please help me to solve this probability question.....
is it the way to do this question... P(x=1 and x=3)=1/4 x 1/4 x 1/4

=0.75 - September 21st 2012, 11:41 PMchiroRe: please help me to solve this probability question.....
No. It looks like you're just guessing: if you don't understand a particular thing, it's best if you just say what those things are rather than guess because you can't really do what you're doing in mathematics.

- September 22nd 2012, 07:47 AMsharmalaRe: please help me to solve this probability question.....
Sorry...I really can't solve this question..