SOLUTION: An urn contains 3 white balls and 7 red balls. A second urn contains 8 white balls and 2 red balls. An urn is selected, and the probability of selecting the first urn is 0.2. A bal
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-> SOLUTION: An urn contains 3 white balls and 7 red balls. A second urn contains 8 white balls and 2 red balls. An urn is selected, and the probability of selecting the first urn is 0.2. A bal
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Question 1203834: An urn contains 3 white balls and 7 red balls. A second urn contains 8 white balls and 2 red balls. An urn is selected, and the probability of selecting the first urn is 0.2. A ball is drawn from the selected urn and replaced. Then another ball is drawn and replaced from the same urn. If both balls are white, what are the following probabilities? (Round your answers to three decimal places.)
(a) the probability that the urn selected was the first one
(b) the probability that the urn selected was the second one Found 2 solutions by ikleyn, math_tutor2020:Answer by ikleyn(52781) (Show Source):
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An urn contains 3 white balls and 7 red balls. A second urn contains 8 white balls and 2 red balls.
An urn is selected, and the probability of selecting the first urn is 0.2.
A ball is drawn from the selected urn and replaced.
Then another ball is drawn and replaced from the same urn.
If both balls are white, what are the following probabilities?
(Round your answers to three decimal places.)
(a) the probability that the urn selected was the first one
(b) the probability that the urn selected was the second one
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Both questions (both parts) are about calculating a conditional probability.
(a) This conditional probability is the fraction, whose denominator is the probability
to get white balls in the first and the second drawn from ANY of the TWO urns,
while the numerator is the probability to get white balls in the first and the second drawn
from the FIRST urn.
Keeping it in mind, we write the formula immediately
P = = = 0.034 (rounded as requested). ANSWER
(b) This conditional probability is the fraction, whose denominator is the probability
to get white balls in the first and the second drawn from ANY of the TWO urns,
while the numerator is the probability to get white balls in the first and the second drawn
from the SECOND urn.
Keeping it in mind, we write the formula immediately
P = = = 0.966 (rounded as requested). ANSWER
A = 1st urn
B = 2nd urn
W = ball is white
R = ball is red
P(A) = 0.2 = 1/5
P(B) = 0.8 = 4/5
P(WW given A) = probability of two white balls, with replacement, given urn A was selected
P(WW given A) = P(W given A)*P(W given A)
P(WW given A) = (3/10)*(3/10)
P(WW given A) = 9/100
P(WW given B) = P(W given B)*P(W given B)
P(WW given B) = (8/10)*(8/10)
P(WW given B) = 64/100
Apply the law of total probability
P(WW) = probability of two white balls, with replacement
P(WW) = P(WW and A) + P(WW and B)
P(WW) = P(WW given A)*P(A) + P(WW given B)*P(B)
P(WW) = (9/100)*(1/5) + (64/100)*(4/5)
P(WW) = 53/100
Then apply the conditional probability formula
P(A given WW) = P(A and WW)/P(WW)
P(A given WW) = P(A and WW) divide P(WW)
P(A given WW) = P(WW given A)*P(A) divide P(WW)
P(A given WW) = (9/100)*(1/5) divide (53/100)
P(A given WW) = (9/100)*(1/5) * (100/53)
P(A given WW) = 9/265
P(A given WW) = 0.034 approximately
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Work Shown for part (b)
Since the options are either "urn 1" or "urn 2", this allows us to subtract the result of part (a) from 1.
P(B given WW) = 1 - P(A given WW)
P(B given WW) = 1 - 0.034
P(B given WW) = 0.966 approximately
Or you could follow a very similar template as done with part (a).
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