00:01
In this problem, we're told that an urn has 12 balls, four of which are white.
00:07
Three players, a, b, and c, are going to draw from the urn, and we want to know the probability for each player that they'll be the first one to draw a white ball.
00:18
We also want to answer this question with replacement between drawings and without replacement between drawings.
00:27
So first, let's ask ourselves, what is the problem? ability that player a, the first player to draw, will be the first player to draw a white ball, assuming that there is a replacement of the ball after each drawing.
00:43
Okay, so right away, with player a's very first drawing, they have a one -third chance of drawing a white ball, since there are four white balls and 12 balls total.
00:53
After, if they fail to draw a white ball, well, that's a two -thirds chance right there.
01:02
And then, each player b and c will also have to fail in order for it to get passed back to player a.
01:13
So player b gets a one -third chance, player c gets a one -third chance.
01:19
So either player a draws the white ball at the very first drawing.
01:23
That's one -third possibility, or in a two -thirds possibility, it's still possible they could draw a white ball if both players b and c fail.
01:34
But then player a succeeds again, succeeds this time, i mean on the second time.
01:45
In that case, there is again two thirds times two thirds times two thirds possibility that will come back around but that they'll succeed the third time and so on and so forth.
02:00
So as we can see there's a pattern here.
02:04
This pattern is actually the same as this pattern.
02:09
Once the ball, once the play has been passed around all three players, the odds of player a being the first to draw the white ball are the same as if they had been when the game had started out.
02:28
So in order to solve for p of a, we're going to observe that this value is the same as this value, and we're just going to say this is equal to p of a.
02:43
And then we're going to simplify this, and we're going to get.
02:48
So assuming that we replace the balls, the probability that a will be the first to draw a white ball is 9 out of 19.
02:56
Now we want to know that under replacement, what is the probability of b and c? well, the probability that b is the first to draw a white ball is actually kind of similar to the probability of a.
03:10
First, a has to fail to draw the white ball first, but after that we can just treat the game as though b is the first player, because from this point on, the play will pass around in a circle, around the three players indefinitely.
03:26
But now we're starting with player b.
03:29
So this is actually just two -thirds of player a, which is equal to six out of 19.
03:36
And it's a similar reasoning for player c, except the ball has to, the play has to pass between players a and b first.
03:45
So that's a four out of nine chance that it passes both of them.
03:49
But at that point, we might as well treat it as the player c is the first player.
03:53
And so we get four out of nine times nine out of 19 is four at 19.
04:02
And so just to check that this all adds up, we want to check that p of a plus p of b plus p of c equals 1, and that's true.
04:13
So that's just a bit of a safety check to make sure that our answer is plausible.
04:18
It doesn't prove that our answer is right.
04:20
But if we had gotten any other value than one, then we wouldn't have known our answer...