One bag contains 4 white balls and 6 black balls. Another bag contains 8 white balls and 2 black balls. A coin tossed to select a bad, then a ball is randomly selected from that bag.
Suppose white ball was drawn. What is the probability that it came from the first bag?
Since a coin toss is used to select either bag1 or bag2
the prob that bag1 is choses is 1/2
prob that a white ball is choses from bag1 = 4/10 = 2/5
so prob of your 'event' = (1/2)(2/5) = 1/5
but the answer in my book says 1/3... Maybe its wrong?
PR=BROWN
@ the man - yes your book is wrong when it says 1/3, which is just totaling white balls in bag 1 (4) and bag 2 (8) to get 12, of which the 1st 4 represent 1/3.
this can can easily be shown to be wrong, if you make the 1st bag 4 white and 600 black instead, your book would still say 1/3, but the probability a white ball from bag 1 would in fact be much, much lower.
incidentally, 1/5 is close, but also wrong. the correct answer is 1 / 5.5.
To find the probability that the white ball came from the first bag, we can apply Bayes' theorem. Bayes' theorem relates conditional probabilities to marginal probabilities. In this case, we need to find the probability of drawing a white ball given that it came from the first bag (P(W|B1)), and the probability of drawing a white ball from either bag (P(W)).
Let's define the events:
B1: Selecting the first bag
B2: Selecting the second bag
W: Drawing a white ball
We want to calculate P(B1|W), which is the probability of selecting the first bag given that a white ball was drawn. According to Bayes' theorem, this can be calculated as:
P(B1|W) = (P(W|B1) * P(B1)) / P(W)
P(W|B1) is the probability of drawing a white ball given that the first bag was selected. There are 4 white balls and 10 balls in total in the first bag, so P(W|B1) = 4/10 = 2/5.
P(B1) is the probability of selecting the first bag without any additional information. Since there are 2 bags in total and each bag is equally likely to be selected, P(B1) = 1/2.
P(W) is the probability of drawing a white ball, regardless of which bag it came from. It can be calculated using the law of total probability:
P(W) = P(W|B1) * P(B1) + P(W|B2) * P(B2)
P(W|B2) is the probability of drawing a white ball given that the second bag was selected. There are 8 white balls and 10 balls in total in the second bag, so P(W|B2) = 8/10 = 4/5.
P(B2) is the probability of selecting the second bag without any additional information. Since there are 2 bags in total and each bag is equally likely to be selected, P(B2) = 1/2.
Plugging in the values, we get:
P(W) = (2/5 * 1/2) + (4/5 * 1/2) = 1/5 + 2/5 = 3/5
Now we can calculate P(B1|W):
P(B1|W) = (2/5 * 1/2) / (3/5) = 2/10 / 3/5 = (2/10) * (5/3) = 1/3
Therefore, the probability that the white ball came from the first bag is 1/3.
Box X contains 8 oranges of which 3 are Bad and Box Y contains 5 pears of which 2 are bad. A fruit is drawn from each box.
a) What is the probability that both are not bad?
b) What is the probability that one item is bad and one is not?
c) If one fruit is a bad and one is not what is the probability that the bad fruit came from box X