# probability that if a black ball is obtained in the second draw a black ball was initially taken out of the box

A box contains 2 black and 1 white ball. A ball is taken out of the box, its colour noted and then the ball is returned to the box together with two more balls of the same colour. Later, balls are drawn from the box one by one until a black ball is obtained.Compute the probability that if a black ball is obtained in the second draw a black ball was initially taken out of the box and returned to it with two more black balls.

Here's how I tackled the problem. Could you see whether i'm right and hint me if i'm wrong?

We have 2 cases, 1st is where we take black ball and return it with two more black balls and second is where we take white ball and return it with two more white balls. So here we go for first case probability:

P(obtaining black ball from second draw)= (2/6)(4/5)(2/3)=4/15 since 2 out of 6 choices for white ball and then 4/5 choices for black ball and 2/3 is probability we obtain black ball from initial draw.

So let the above probability denote B2. Bi would denote event of taking out black ball from initial set of balls.P(Bi)=2/3

Now, I calculate P(Bi | B2)= P(Bi and B2)/P(B2)= P(B2|Bi)P(Bi)/P(B2)= (4/15)(2/3)/P(drawing from second time when white initially + -||- black initially)= (4/15)(2/3)/(4/15+3/10)= 4/17

If it's incorrect could you please hint

• To be clear, the second draw is the first draw that occurs after you add to balls to the box, right? Commented Aug 6, 2019 at 20:17
• @RobertShore no, there is one initial draw= draw 0. Then draw 1,2,3.. are draws once we obtained 5 or 6 balls
– user634512
Commented Aug 6, 2019 at 20:19
• When you say "balls are drawn from the box one by one," is that without replacement? Commented Aug 6, 2019 at 20:21
• @RobertShore yes, I mean to be honest that's what I think the exercise wants me to do. Although not 100% sure but i think yes
– user634512
Commented Aug 6, 2019 at 20:23
• After the replacement, there are only $5$ balls in the box, not $6$. Commented Aug 6, 2019 at 20:34

Let's say balls $$1$$ and $$2$$ are black and ball $$3$$ is white. Then you have the following possibilities:
In $$20$$ of $$30$$ trials, your "zeroth" draw will be black. In $$16$$ of those $$20$$ trials, your first draw will be black and there won't be a second draw. In all of the remaining $$4$$ trials, your second draw will be black.
In $$10$$ of $$30$$ trials, your "zeroth" draw will be white. In $$4$$ of those $$10$$ trials, your first draw will be black and there won't be a second draw. In $$3$$ of the remaining $$6$$ trials, your second ball will be black.
Each of these trials occurs with equal probability. There are seven trials that result in a black ball on the second (post-replacement) draw. Of those, your initial draw was black in four of them. So the probability you want is $$\frac 47$$.