Question
Suppose we have an urn with 10 white balls and 6 red balls, and we create an experiment by randomly selecting 6 balls from the urn.
a) What is the probability that the sample has the same number of white balls as red balls?
b) What is the probability that there are more red balls than white balls in our sample?
c) What is the probability that the sample has at least 5 white balls?
a) What is the probability that the sample has the same number of white balls as red balls?
b) What is the probability that there are more red balls than white balls in our sample?
c) What is the probability that the sample has at least 5 white balls?
Answers
Assuming that the balls are replaced, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
a) (6/16)^3 * (10/16)^3 = ?
If they are not replaced,
a) 6/16 * 5/15 * 4/14 * 10/13 * 9/12 * 8/11 = ?
Depending on replacement or not, go through the similar processes with the remaining problems.
b) Here you are considering 4 or 5 red balls. "More" indicates that there has to be at least one white ball. (Either-or probabilities are found by adding the individual probabilities.)
c) Here you are considering 5 or 6 white balls
a) (6/16)^3 * (10/16)^3 = ?
If they are not replaced,
a) 6/16 * 5/15 * 4/14 * 10/13 * 9/12 * 8/11 = ?
Depending on replacement or not, go through the similar processes with the remaining problems.
b) Here you are considering 4 or 5 red balls. "More" indicates that there has to be at least one white ball. (Either-or probabilities are found by adding the individual probabilities.)
c) Here you are considering 5 or 6 white balls
a). 300/1001
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