I am trying to do the task of finding the probability of taking white ball of the three boxes. Task text is -
In the first box, there are $4$ white and $1$ black ball.
In the second - $3$ white and $2$ black balls.
And in the third - $1$ white and $3$ black balls.
From the first two boxes, one ball is being taken from each, and they are put in the third box. Then they are shaken and one ball is being taken from the third box.
Find a probability of taking white ball of the third box.
There are two things, which are causing difficulties for me.
1) How to handle different situations (If taken balls are all black or all white etc.)
2) Which formula to use for getting the right probability. I have tried to use this formula $P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)$ And after all calculations, I got the probability of $93.6$%, which is a bit too much in my opinion.
And I am confused, if I should use on third box ($P(C$)) total of $4$ balls or $6$ balls. Thanks ;)
Use the law of total probability. Let $E$ be the probability thatwe take a white ball out of the third box. In the first stage of the experiment we either draw $WW$, $BB$, $WB$ or $BW$ where the first letter corresponds to the draw from the first box and the second letter corresponds to the draw from the second box, $W$ is white and $B$ is black. Then $$ P(E)=P(E\mid WW)P(WW)+P(E\mid BB)P(BB)+P(E\mid WB)P(WB)+P(E\mid BW)P(BW). $$ These probabilities can be computed as follows. For example $$ P(WW)=\frac{4}{5}\times \frac{3}{5} $$ and $$ P(E\mid WW)=\frac{3}{6} $$ since after adding the white balls there are $6$ balls in total in the third box of which $3$ are white. Other probabilities can be computed similarly.