I am struggling with the math conditional probability task:
There are two boxes. One has 14 white and 32 black balls, another has 23 white and 43 black balls. Choosing a box randomly, we choose a white ball. What is a probability, that from the same box we will choose a white ball again?
My solution:
In general, there are 46 balls in the first box, and 66 box in the second one. There is a condition, that we already chose a white ball and we gonna choose from the same box white ball.
So, to choose a ball from the first box again, we have a probability:
$$\frac{14-1}{46-1} \cdot \frac{1}{2}$$
and from the second box:
$$\frac{23-1}{66-1} \cdot \frac{1}{2}$$
I know, that If I sum these two probabilities the result will be wrong because I did not specify a condition. I believe I need to use Bayes' theorem.
Thus:
$$P(A \mid B) = \frac{P(A~\text{and}~B)}{P(B)}$$
A is event of choosing a white ball again from the same box knowing that B is true (B = firstly we chose a white ball from the random box)
$$P(A~\text{and}~B) = P(A) \cdot P(B)$$
Then we have:
$$\frac{1}{2} \cdot \frac{P(\text{Probability of choosing a white ball})}{P(\text{Probability of choosing a white ball in general})}$$
I am getting a wrong result and I am struggling with the correctness of my solution
Best Answer
We wish to find the probability that a second white ball is selected from the same box given that a white ball has been selected from one of the boxes.
Let $F$ be the probability that a white ball is selected from one of the boxes. Then $$\Pr(F) = \frac{1}{2} \cdot \frac{14}{46} + \frac{1}{2} \cdot \frac{23}{66}$$
Let $E$ be the event that a second white ball is selected from the same box. Then $$\Pr(E \cap F) = \frac{1}{2} \cdot \frac{14}{46} \cdot \frac{13}{45} + \frac{1}{2} \cdot \frac{23}{66} \cdot \frac{22}{65}$$
Hence, \begin{align*} \Pr(E \mid F) & = \frac{\Pr(E \cap F)}{\Pr(F)}\\ & = \frac{\frac{1}{2} \cdot \frac{14}{46} \cdot \frac{13}{45} + \frac{1}{2} \cdot \frac{23}{66} \cdot \frac{22}{65}}{\frac{1}{2} \cdot \frac{14}{46} + \frac{1}{2} \cdot \frac{23}{66}} \end{align*}