Question

We've got 4 boxes and 4 balls in each. In box number k there are k black balls and (4-k) white balls. We choose 1 box at random (the same probability for each). From chosen box we draw black ball. What's the probabilty to draw black ball again?

Answer 1

we can do this

Answer 2

we know we have chosen a black ball for the first one
it came from the the first box with probability \(\frac{1}{4}\times \frac{1}{4}\)
second with probability \(\frac{1}{4}\times \frac{1}{2}\)
third with probability \(\frac{1}{4}\times \frac{3}{4}\)
forth one with probability \(\frac{1}{4}\times 1\)

Answer 3

now what is the probability that we pick a second ball black ball given that the first ball is black?
if we pick from the first one, with probability \(\frac{1}{4}\) then the probability we get a black ball is zero, because there are none left

if we pick from the second with probability \(\frac{1}{4}\) then the probability it is black is \(\frac{1}{3}\) since there is one black ball and three total

if we pick from the third with probability \(\frac{1}{4}\) then the probability you get a black ball is \(\frac{2}{3}\) because there are two black left, and 3 total, and if you pick from the fourth with probability \(\frac{1}{4}\) then you are sure to pick a black ball because there are nothing but black balls in it

Answer 4

so your answer is, the probabilty the second is black given the first on is, is
\[\frac{1}{4}(0+\frac{1}{3}+\frac{1}{3}+1)\]