# probability of drawn a ball from same bag without replacement

Problem 1: 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?

Answer 1/2: There is two bags.One containing black balls and other contain 1 black and 1 white ball. If I drawn 1st ball black then the 2nd ball is also black if I chose the bag which contain 2 black ball. So to get 2nd ball also black probability is 1/2.

Answer 3/4: Total ball 4. 1st one drawn. So next one when drawn probability of getting black ball is 3/4 compare to remaining ball.

Which one is correct?

• Define $A$ as the event that you pick the bag with two black balls. Define $B$ as the event that the first ball you saw was black. So you want to compute $P[A|B]$, no? Do you know how to use/switch conditionings? – Michael Nov 14 '18 at 22:01
• What is switch conditionings? – RealCoder Nov 14 '18 at 22:06
• Switch conditioning. Sometimes called Baye's rule. It gives you $P[A|B]$ in terms of $P[B|A]$ and some other probabiltiies. It just applies the definition of conditional probability. – Michael Nov 14 '18 at 22:07
• oh okk.. I know baye's rule. – RealCoder Nov 14 '18 at 22:08
• Okay so you should be open to the possibility that neither of your guesses is correct. Use Baye's rule. – Michael Nov 14 '18 at 22:09

When given that you have drawn a black ball, the ball that you drew is more likely to have come from the bag with more black balls.   So the answer is not $$1/2$$.
Now, as the second answer correctly argues, each individual ball had been equally likely to be drawn.   However, on considering how many among all the black balls come from the bag with a another black ball, we find the answer is not $$3/4$$.