Posted by Aubrey on Monday, November 20, 2006 at 7:22pm.
Bowl one contains 3 black and 2 white balls. Bowl two contains 4 black and 5 white balls.
Experiment: A ball is randomly selected from bowl one, its color noted, and it is placed in bowl two. Then a ball is randomly selected from bowl two.
a.) What is the probability that the second ball selected is white?
b.)If the second ball selected is white, what is the probability that the first ball selected was black?
c.) Given: If both picks black, I win $6, if there is one pick of each color, I lose $1, and if both picks are white I lose $5. What are my expected winnings?
You have to sum the probabilities of all ways that lead to a particular outcome to obtain the probability of that outcome.
In this case there are two possibilities: The black ball goes to bowl 2 or the white ball goes there.
Suppose the black ball goes there. The probability of that happening is 3/5. The probability that you then draw the white ball (i.e. the conditional probability that the white ball is drawn given that the black ball placed in bowl two) is 1/2.
The overall probability of this particular way that leads to the white ball being selected is thus 3/5*1/2 = 3/10.
Note that if you multiply the conditional probability of some event X happening given that some other event Y has happened and multiply that with the probability that Y would happen in the first place, you get the joint probability of X and Y happening.
In this case X stands for the white ball being selected in bowl 2 and Y stands for the balck ball being transferred to bowl 2 from bowl 1.
The probability that the white ball is transferred is 2/5. The probability that you draw the white ball from bowl 2 given that the white ball is transferred is 6/10. Therefore the probability that the white ball is srawn and the white ball is transferred is 2/5 * 6/10 = 6/25
The probability of drawing the white ball, irrespective of which ball was transferred is thus:
3/10 + 6/25 = 27/50
THANK YOU!
It's a pleasure to help! In part b) you must divide the probability of 3/10 by 27/50, so you get 15/27.
It's again the same rule. The joint probability of X and Y is the probability of Y happening multiplied by the probability of X happening (or that it has happened) given that Y has happened. In this case you take Y to be the event that the white ball is drawn from the second bowl.
The joint probability that the black ball is transferred and the white ball is drawn was 3/10 as we already evaluated in part a). So, we must divide this joint probability by the probability of the white ball being drawn (which we found to be 27/50) to obtain the conditional proability that it was the black ball that was transferred given that the white ball was drawn.

Finite  Bob, Wednesday, January 21, 2015 at 11:08pm
4/6
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