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標題:
有冇人讀過stats, 有野想問..:(
發問:
1a) A bag contains 4 red counters and 6 green counters. Four counters are drawn at random from the bag, without replacement. Calculate the probability thati) All the counters are green;ii) at least one counter of each colour is drawn;iii) at least two green counters are drawn; given that at least one of... 顯示更多 1a) A bag contains 4 red counters and 6 green counters. Four counters are drawn at random from the bag, without replacement. Calculate the probability that i) All the counters are green; ii) at least one counter of each colour is drawn; iii) at least two green counters are drawn; given that at least one of each colour is drawn. b) Are the events "at least two green counters are drawn" and "at least one counter of each colour is drawn" independent? Explain briefly. 2) A coin and a six-faced die are thrown simultaneously. The random variable X is defined as follows: "If the coin shows a head, then X is the score on the die. If the coin shows tail, then X is twice the score on the die." a) Find the expected value of X. b) Find the variance of X If the experiment is repeated and the sum of the two values obtained for X is denoted b Y. c) Find P (Y=4) and E(Y) 唔該哂:-(
最佳解答:
X Pr(X) X * Pr(X) X2 * Pr(X) 1 0.0833 0.0833 0.0833 2 0.1667 0.3333 0.6667 3 0.0833 0.2500 0.7500 4 0.1667 0.6667 2.6667 5 0.0833 0.4167 2.0833 6 0.1667 1.0000 6.0000 8 0.0833 0.6667 5.3333 10 0.0833 0.8333 8.3333 12 0.0833 1.0000 12.0000 S 5.2500 37.9167 Y Pr(Y) Y * Pr(Y) 2 0.0069 0.0139 3 0.0278 0.0833 4 0.0417 0.1667 5 0.0556 0.2778 6 0.0764 0.4583 7 0.0833 0.5833 8 0.0972 0.7778 9 0.0694 0.6250 10 0.0903 0.9028 11 0.0556 0.6111 12 0.0833 1.0000 13 0.0417 0.5417 14 0.0833 1.1667 15 0.0278 0.4167 16 0.0625 1.0000 17 0.0139 0.2361 18 0.0417 0.7500 20 0.0208 0.4167 22 0.0139 0.3056 24 0.0069 0.1667 S 10.5000 2010-04-16 09:50:07 補充： 1ai Pr = 6/10 x 5/9 x 4/8 x 3/7 = 1/14 = 0.0714 2010-04-16 09:50:18 補充： 1aii Pr = 1 – Pr(all green) – Pr(all red) = 1 – 1/14 – (4/10 x 3/9 x 2/8 x 1/7) = 1 – 1/14 – 1/210 = 97/105 = 0.9238 2010-04-16 09:50:21 補充： Pr(at least 2 green | at least 1 green) = Pr(at least 2 green) / Pr(at least 1 green) = [1 – Pr(all red) – Pr(1 red)] / [1 – Pr(all red)] = [1 – 1/210 – (4/10 x 6/9 x 5/8 x 4/7) x 4] / (1 – 1/210) = (1 – 1/210 – 2/21) / (1 – 1/210) = (9/10) / (209/210) = 189/209 = 0.9043 2010-04-16 09:50:34 補充： 2a: E(X) = 5.25 2b: Var(X) = E(X2) – E2(X) = 37.9167 – 5.252 = 10.3542 2010-04-16 09:50:48 補充： X1X2YPr(Y) 1340.0069 2240.0278 3140.0069 ?0.0417 Pr(Y=4) = 0.0417 2010-04-16 09:51:02 補充： E(Y) = 10.5
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