# Thread: Conditional Probability Problem.

1. ## Conditional Probability Problem.

Hi, another actuary problem here:

Two bowls each contain 5 black and 5 white balls. A ball is chosen at random from bowl 1 and put into bowl 2. A ball is then chosen at random from bowl 2 and put into bowl 1. Find the probability that bowl 1 still has 5 black and 5 white balls.

So the actuary answer had more algebra, I will give their solution if I am wrong, but it seems I am right. I thought of the problem like this:

We denote P[A] and as the white ball transferring from bowl 1 to bowl 2. So then we denote P[A'] as black ball transferring from bowl 1 to bowl 2.

Also we denote P[B] as the white ball going back to bowl 1. We then denote P[B'] As the black ball going back to bowl 1.

P[ B | A]= 6/11 <--- this is the probability of white ball going back to bowl 1, given that the white bowl went into bowl 2.
Similiarily, we then know that P [B' | A'] = 6/11 as well, because it is the same event happening just with different colored ball.

Then we also know that P[ B' | A ] = P[ B | A'] = 5/11

So the probability in question is find probability : P[ B | A] or... P[ B' | A'], which = 6/11. And that is my final answer. IS this the most efficient way to solve the problem? To me it makes the most sense.

2. ## Re: Conditional Probability Problem.

How about this: whatever color ball moves from bowl 1 to 2 is the same color as the ball that must move back. There will be 6/11 of that color ball. That's just one step. Is that more efficient?

Yes, your solution is correct.

3. ## Re: Conditional Probability Problem. Originally Posted by math951 Hi, another actuary problem here:
Two bowls each contain 5 black and 5 white balls. A ball is chosen at random from bowl 1 and put into bowl 2. A ball is then chosen at random from bowl 2 and put into bowl 1. Find the probability that bowl 1 still has 5 black and 5 white balls.

So the probability in question is find probability : P[ B | A] or... P[ B' | A'], which = 6/11. And that is my final answer. IS this the most efficient way to solve the problem? To me it makes the most sense.
I am adding a reply only to show another way with different notation.
Using $W_1~\&~W_2$ for the first and second balls are white.
$\mathcal{P}(W_1\cap W_2)=\mathcal{P}(W_2|W_1)\cdot\mathcal{P}(W_1)= \dfrac {6}{11}\cdot\dfrac{5}{10}=\dfrac{3}{11}$ then double that to get your answer.