Probability

[匿名]

2010-12-23 6:52 am

There are 4 cards on the table : '1', '2', '3' and '4'. A dice is tossed and a card is removed from the table if their numbers are the same. Find:
(1) Probability that card '1' is removed before card '2'.
(2) After tossing for 3 times, probability that card '1' and '2' are still on the table.
(3) After tossing for 3 times, probability that card '1' and '2' have been removed.

回答 (1)

用戶9112

2010-12-23 7:52 am

✔ 最佳答案

(1) Since 1 or 2 has same probability of being removed first (by symmetry),Required probability =0.5(2) Probability that none of the 2 is removed during a toss = 4/6 = 2/3After 3 tosses, the probability for both number to remain = (2/3)^3 = 8/27(3) The sequence of removal of 1 and 2 can be 12X, 1X2, 21X, 2X1, X12 or X21The related probability = (1/6)(1/6)(6/6) + (1/6)(5/6)(1/6) +(1/6)(1/6)(6/6)+(1/6)(5/6)(1/6)+(4/6)(1/6)(1/6) + (4/6)(1/6)(1/6)=5/36

2010-12-23 20:07:05 補充：
Alternative solution for (3):
Pr that 1 will remain after 3 trials = (5/6)^3 = 125/216
Pr that 2 will remain after 3 trails = 125/216
Pr that 1 or 2 remain = Pr(1) + Pr(2) - Pr(1 and 2) = 125/216 + 125/ 216 - 8/27 = 186 /216
Pr that both 1 and 2 are removed = 1 - 186/216 = 30/216 = 5/36

👍 0 👎 0