SOLUTION: Consider an urn that contains 4 blue and 7 red balls. First one ball is selected, and then a second ball is selected without replacement.
(a) What is the probability that at lea
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-> SOLUTION: Consider an urn that contains 4 blue and 7 red balls. First one ball is selected, and then a second ball is selected without replacement.
(a) What is the probability that at lea
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Question 1131560: Consider an urn that contains 4 blue and 7 red balls. First one ball is selected, and then a second ball is selected without replacement.
(a) What is the probability that at least 1 red ball is chosen?
(b) What is the probability that the second ball chosen is red, given the that the first ball chosen was red?
(c) What is the probability that the second ball chosen is red, given that the first ball chosen was not red?
(a) P(at least one red ball) = 1-P(no red ball) = P(both balls blue)
(b) P(second ball red | first ball red)
By the definition of conditional probability, this is equal to
P(first red AND second red)
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P(first red)
I, as well as many students, find the formal definition of conditional probability, in the form of a formula for calculating it, somewhat confusing. For me it is easier to understand this way:
"...given that the first ball is red" means the only cases we are considering are those in which the first ball is red; that means the denominator of the probability fraction, representing the "sample space", is only the probability of getting red with the first ball: P(first red). Then the numerator is the probability that we ALSO get a red on the second ball: P(first red AND second red).
(c) P(second ball red | first ball blue)
This is
P(first blue AND second red)
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P(first blue)
