Two boxes, the first contains 3 red & 2 white balls. The second box contains 4 red & 7 white balls.
A fair die is tossed. If the outcome is divisible by 2, a ball is drawn from the first box, while a ball is drawn from the second box if the outcome is not divisible by 2.
Find the probability that a randomly chosen ball will be white and it comes from the first box.
Attempt at solution:
$P(drawing from first box)=\frac{3}{6}=\frac{1}{2}$
$P(drawing a white ball) = \frac{9}{16}$
$P(white ball \cap first box) = $ $P(white ball | first box) \times P(First box)= \frac{2}{5}\frac{1}{2} = \frac{1}{5}$,
My friend says that the required probability is not $P( White ball \cap first box)$ but instead it's $P(White ball | first box)$, which is it?
You want the probability of choosing a white ball from the first box. You don't want the probability of choosing a white ball GIVEN that you pick the first box. Nothing is given to you at this point so your friend would be wrong. Multiply the probability of choosing the first box by the probability of choosing a white ball given that you choose the first box.
$P(first box ) \times P( white ball | first box )= \frac{1}{2}\frac{2}{5} = \frac{1}{5}$ would be correct.