1.Suppose that eight families live on the same floor of a building where five own their homes and three of them rent. If four families are to be chosen at random without replacement to form a sample and if X is a random variable representing the number of homeowners in the sample,
a.what are the possible values of X?
b. What is the probability of getting two homeowners in the sample ?
c. what is the probability of getting fewer than two families who rent their home in the sample?
Probability - Urgent!!!
2008-05-26 6:32 pm
回答 (2)
2008-05-27 11:11 pm
a. 1, 2, 3 or 4
b. (4C2)*(5/8)*(4/8)*(3/8)*(2/8) = 45/256
c. (4C0)*(5/8)*(4/8)*(3/8)*(2/8) + (4C1)*(3/8)*(5/8)*(4/8)*(3/8) = 105/512
2008-05-28 08:47:11 補充：
Sorry, 計錯, 更正如下:
a. 1, 2, 3 or 4
b. (4C2)*(5/8)*(4/7)*(3/6)*(2/5) = 3/7
c. (4C0)*(5/8)*(4/7)*(3/6)*(2/5) + (4C1)*(3/8)*(5/7)*(4/6)*(3/5) = 1/2
b. (4C2)*(5/8)*(4/8)*(3/8)*(2/8) = 45/256
c. (4C0)*(5/8)*(4/8)*(3/8)*(2/8) + (4C1)*(3/8)*(5/8)*(4/8)*(3/8) = 105/512
2008-05-28 08:47:11 補充：
Sorry, 計錯, 更正如下:
a. 1, 2, 3 or 4
b. (4C2)*(5/8)*(4/7)*(3/6)*(2/5) = 3/7
c. (4C0)*(5/8)*(4/7)*(3/6)*(2/5) + (4C1)*(3/8)*(5/7)*(4/6)*(3/5) = 1/2
2008-05-26 8:56 pm
a. X = 1, 2, 3 or 4
b. P(X = 2) = (4C2) * (5/8)^2 * (3/8)^2 = 675/2048
c. P(X >= 3) = (4C3) * (5/8)^3 * (3/8)^1 + (4C4) * (5/8)^4 * (3/8)^0
2008-05-26 14:18:18 補充：
c. = 2125/4096
b. P(X = 2) = (4C2) * (5/8)^2 * (3/8)^2 = 675/2048
c. P(X >= 3) = (4C3) * (5/8)^3 * (3/8)^1 + (4C4) * (5/8)^4 * (3/8)^0
2008-05-26 14:18:18 補充：
c. = 2125/4096
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