# Struggling with conditional probability question

• April 10th 2011, 11:25 AM
angypangy
Struggling with conditional probability question
Hello

The question says:

A bag P contains three red balls. A second bag Q contains two red balls and three black balls.
(i) A bag is chosen at random and one ball is withdrawn. Find the probability that this ball is red.

OK, using a tree diagram I correctly worked out the answer to be 7/10. no problem there.

But the rest of the question.

This ball remains outside the bag.
(ii) A bag is again chosen at random (it is not known whether this is the same bag as before or not) and one ball is withdrawn. Find the joint probability that both this ball and the one previously withdrawn are red.

To answer this one I calculated that if ball chosen from bag P then it could only be red - 1/2 chance of that. For the Q bag would be a 1/5 chance of getting a red ball. (Ah just seen a problem with that logic - how do you know there are still 5 balls in Q???

OK, how would you lay this one out? Would you use a tree diagram? Some help would be much appreciated. I have studied conditional probability but just stumped on this question.

I didn't go on to the next part of the question:
(iii) If they are both red, what is the probability that bag P was used on both occasions?

Angus
• April 10th 2011, 11:58 AM
Plato
Quote:

Originally Posted by angypangy
The question says:
A bag P contains three red balls. A second bag Q contains two red balls and three black balls.
(i) A bag is chosen at random and one ball is withdrawn. Find the probability that this ball is red.

$\mathcal{P}(R)=\mathcal{P}(R|P)\mathcal{P}(P)+\mat hcal{P}(R|Q)\mathcal{P}(Q)=1\cdot\frac{1}{2}+\frac {2}{5}\cdot\frac{1}{2}$
• April 10th 2011, 01:05 PM
angypangy
Yes, thanks but I got that bit. It was question (ii) I was stuck on.
• April 10th 2011, 01:16 PM
Plato
Quote:

Originally Posted by angypangy
Yes, thanks but I got that bit. It was question (ii) I was stuck on.

$\mathcal{P}(R_1R_2)=\mathcal{P}(R_2|R_1)\mathcal{P }(R_1)$.

You know $\mathcal{P}(R_1)$ go about finding $\mathcal{P}(R_2|R_1)$ in exactly the same using one less red ball.