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?
1 5/9
2.3/6
3.12
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 ?
a) To find the probability that the second ball selected is white, we can use the principle of conditional probability. The probability of selecting a white ball from bowl two depends on the outcome of the first selection from bowl one.
Let's break it down step by step:
1. Calculate the probability of selecting a white ball from bowl one:
- Bowl one contains 3 black balls and 2 white balls, so the probability of selecting a white ball from bowl one is 2/5.
2. Calculate the probability of selecting a white ball from bowl two, given that a white ball was selected from bowl one:
- After the first selection, the contents of bowl two have changed. Bowl one lost one white ball, and bowl two gained one white ball.
- Bowl two now contains 4 black balls and 6 white balls.
- The probability of selecting a white ball from bowl two, given that a white ball was selected from bowl one, is 6/10.
3. Multiply the probabilities from steps 1 and 2 to get the overall probability:
- (2/5) * (6/10) = 12/50 = 6/25
- The probability that the second ball selected is white is 6/25.
Therefore, the correct answer to a) is 6/25, not 54%.
b) To find the probability that the first ball selected was black, given that the second ball selected is white, we can use Bayes' theorem.
Let's break it down step by step:
1. Calculate the probability of selecting a black ball from bowl one, when the second ball selected is white:
- The probability of selecting a black ball from bowl one, given that the second ball selected is white, is denoted as P(B|W).
- From part a), we found that the probability that the second ball selected is white is 6/25.
- Now, we need to find the probability that the first ball selected was black and the second ball selected is white (P(BW)).
2. Calculate the probability of selecting a black ball from bowl one and a white ball from bowl two (P(BW)):
- We already know from part a) that the probability of selecting a white ball from bowl two, given that a white ball was selected from bowl one, is 6/10.
- The probability of selecting a black ball from bowl one is 3/5.
- Therefore, P(BW) = (3/5) * (6/10) = 18/50 = 9/25.
3. Use Bayes' theorem to calculate P(B|W):
- Bayes' theorem states that P(B|W) = (P(BW)/P(W)).
- P(W) is the probability of selecting a white ball from bowl two, which is 6/10 (as we calculated in step 2).
- Therefore, P(B|W) = (9/25)/(6/10) = (9/25) * (10/6) = 9/15 = 3/5.
Therefore, the correct answer to b) is 3/5, not 2/3 as mentioned before.
c) To calculate your expected winnings, we need to assign the value of each outcome (win or lose) and multiply it by the corresponding probabilities.
Let's break it down step by step using a probability tree:
1. Calculate the probabilities for each outcome:
- P(B,B) = probability of selecting black on the first draw and black on the second draw = 0.3 (as calculated in the problem statement)
- P(B,W) = probability of selecting black on the first draw and white on the second draw = 0.3 (as calculated in the problem statement)
- P(W,B) = probability of selecting white on the first draw and black on the second draw = 0.16 (as calculated in the problem statement)
- P(W,W) = probability of selecting white on both draws = 0.24 (as calculated in the problem statement)
2. Assign the values for each outcome:
- $6 for P(B,B)
- -$1 for P(B,W)
- -$1 for P(W,B)
- -$5 for P(W,W)
3. Calculate the expected winnings:
- Expected winnings = (P(B,B) * $6) + (P(B,W) * -$1) + (P(W,B) * -$1) + (P(W,W) * -$5)
- Expected winnings = (0.3 * $6) + (0.3 * -$1) + (0.16 * -$1) + (0.24 * -$5)
- Expected winnings = $1.80 - $0.30 - $0.16 - $1.20
- Expected winnings = $0.14
Therefore, the correct answer to c) is $0.14 (14 cents), not 12 as mentioned before.