23.Two cards are drawn without replacement from an ordinary deck of 52 playing cards. If Z is the number of aces obtained in the first draw and W is the total number of aces obtained in both draws, find
a)the joint probability distribution of Z and W;
b)the marginal distribution of Z;
c)the marginal distribution of W;
d)the conditional distribution of W given Z = 1.
數學題,,,,
2007-10-30 6:09 am
回答 (1)
2007-10-30 6:44 am
✔ 最佳答案
There are four cases: XX (with prob = 48/52 * 47/51 = ), at that time, P(Z = 0, W = 0) = 188/221
XA (with prob = 48/52 * 4/51), at that time, P(Z = 0, W = 1) = 16/221
AX (with prob = 4/52 * 48/51), at that time, P(Z = 1, W = 1) = 16/221
AA (with prob = 4/52 * 3/51), at that time, P(Z = 1, W = 2) = 1/221
[Where X denotes non Ace]
That is the joint prob of Z and W, all other cases has zero probability.
2b) P(Z = 0) = 48/52 = 12/13, P(Z = 1) = 4/52 = 1/13, that can be obtained inituitively, or you can marginalize the joint probabilities.
P(Z = 0) = P(Z = 0, W = 0) + P(Z = 0, W = 1) + P(Z = 0, W = 2) = 188/221 + 16/221 + 0 = 12/13
P(Z = 1) = P(Z = 1, W = 0) + P(Z = 1, W = 1) + P(Z = 1, W = 2) = 0 + 16/221 + 1/221 = 1/13
2c) P(W = 0| Z = 1) = 0, P(W = 1| Z = 1) = 48/51 = 16/17, P(W = 2|Z = 1) = 3/51 = 1/17. This can be obtained inituitively, or you can use the definition on conditional probability
P(A = x| B = y) = P(A = x, B = y) | P (B = y), e.g.
P(W = 1| Z = 1) = P(W = 1, Z = 1) / P(Z = 1) = (16/221)/(1/13) = 16/17.
收錄日期: 2021-04-23 17:29:25
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