I was taking Caltech – ML Course and solving problem 1.3 in this link
We have 2 opaque bags, each containing 2 balls. One bag has 2 black balls and
the other has a black ball and a white ball. You pick a bag at random and then pick one of the balls in that bag at random. When you look at the ball, it is black. You now pick the second ball from that same bag. What is the probability that this ball is also black?
I tried to do this way:
- Since we are trying to find the probability of 2nd being ball being black when we already know that first ball is black. This reduces our problem to probability by which we picked the bag with two black balls.
So the answer should be 1/2.
PS: This is not a homework/assignment. This course has been already finished. I am just taking it offline for learning purpose.
Best Answer
Since $2$ of the three black balls are in the $2$-blacks bag, the required probability is $2/3$.
Since intuition can be treacherous, we will in addition do a formal conditional probability calculation.
Let $F$ be the event the first ball is black. The probability of $F$ is $3/4$. For with probability $1/2$, we chose the $2$-blacks bag, in which case the probability of black is $1$, and with probability $1/2$ we chose the mixed bag, in which case the probability of black is $1/2$. Thus $\Pr(F)=(1/2)(1)+(1/2)(1/2)$.
Let $S$ be the event the second ball drawn is black. The probability of $S$ is $1/2$, for $S$ happens precisely if we picked from the $2$-blacks bag.
The event $F\cap S$ happens precisely if $S$ happens. Thus $$\Pr(S|F)=\Pr(F\cap S)/\Pr(F)=\frac{2}{3}.$$