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?
Except for a) Im not sure what the right answers are, Im fairly sure the answers by chris are not right.
These are difficult problems, at a level I did not see until a fairly advanced level mathmatics course in college. The answer to a) by the way is 54%
For c) build a probability tree.
P(B,B) = .6 * .5 = .3
P(B,W) = .6 * .5 = .3
P(W,B) = .4 * .4 = .16
P(W,W) = .4 * .6 = .24
Multiply each of these by the win or loss. So, I believe for c) the expected winnings is 14 cents.
Ok now that I have consulted with my old probability text, I believe I am ready to answer b)
The conditional probability P(B\W)
(probability of black on the first draw given white on the second) is equal to P(BW)/P(xW) (probability of Black on first, White on second) divided by the probability of either black or white on the first and white on the second draw)
P(BW) = .6 * .5 = .3
P(xW) = .54 (from problem a)
So, P(B\W) = .3/.54 = .5555
a container cotains 6 yellow balls , 2pink balls, 3 green balls how can i show in tree diagram and what's the probability choosing 2 green balls ?
posted by Ram