標題:
probaaility
Suppose four people sit down to dinner at a table, and each place is set with 3 pieces of cutlery. In total, there are 4 knives, 4 forks and 4 spoons on the table; but a mischievous butler has allocated these 12 utensils at random. Determine the probability that:a) each person gets one fo each utensil b) one... 顯示更多 Suppose four people sit down to dinner at a table, and each place is set with 3 pieces of cutlery. In total, there are 4 knives, 4 forks and 4 spoons on the table; but a mischievous butler has allocated these 12 utensils at random. Determine the probability that: a) each person gets one fo each utensil b) one person gets all knives, one person gets all forks, and one person gets all spoons. 更新: cccmartin46 part a 答案是對的，但part b不對。 我有答案但唔曉做！ thx~ 更新 2: part b 答案是6.93*10^ -4
最佳解答:
a) P(each person get one of each utensil) =(4!4!4!/12!)*6^4 =0.037402597 Remark: *6 because (f,s,k),(f,k,s),(s,f,k),(s,k,f),(k,s,f)or(k,f,s) also=1set utensil b) P(1 person gets all knives, 1 person gets all forks, and 1 person gets all spoons) =(4/12*3/11*2/10*4/9*3/8*2/7*4/6*3/5*2*4)*4! =(4!4!4!3!/12!)*4! =0.004155844156 Remark:*4! because (fff,sss,kkk,fsk)can be formed 24 diffenece sequences
其他解答:
a) P(each gets one of each utensil ) = 12/12 [1st person can get anything first] x 8/11 [There r 8 remaining pieces of cultery different from the first piece] x 4/10 [There r 4 remaining pieces of cultery different from both the first two pieces] x 9/9 [2nd person can get anything first] x 6/8 [There r 6 remaining pieces of cultery different from the previous piece] x 3/7 [There r 3 remaining pieces of cultery different from the previous two pieces] x 6/6 [3rd person can get anything first] x 4/5 [There r 4 remaining pieces of cultery different from the previous piece] x 2/4 [There r 2 remaining pieces of cultery different from the previous two pieces] [and the last person can get the remaining 3 different pieces in any order] = 1 * 8/11 * 4/10 * 1 * 6/8 * 3/7 * 1 * 4/5 * 2/4 = 72/4235 (or 0.017) b) P(one gets all knives, one gets all forks, and one gets all spoons) = 12/12 [1st person can get anything] x 3/11 [Only 3 same pieces of cultery remaining] x 2/10 [Only 2 same pieces remaining] x 8/9 [8 pieces out of 9 are different from what the 1st person got] x 3/8 [Only 3 same pieces of cultery remaining] x 2/7 [Only 2 same pieces remaining] x 4/6 [4 pieces out of 6 are different from what the 1st and 2nd persons got] x 3/5 [Only 3 same pieces of cultery remaining] x 2/4 [Only 2 same pieces remaining] [and the last person can get the remaining 3 different pieces in any order] = 1 * 3/11 * 2/10 * 8/9 * 3/8 * 2/7 * 4/6 * 3/5 * 2/4 = 2/4235 (or 0.00047)
probaaility
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發問:Suppose four people sit down to dinner at a table, and each place is set with 3 pieces of cutlery. In total, there are 4 knives, 4 forks and 4 spoons on the table; but a mischievous butler has allocated these 12 utensils at random. Determine the probability that:a) each person gets one fo each utensil b) one... 顯示更多 Suppose four people sit down to dinner at a table, and each place is set with 3 pieces of cutlery. In total, there are 4 knives, 4 forks and 4 spoons on the table; but a mischievous butler has allocated these 12 utensils at random. Determine the probability that: a) each person gets one fo each utensil b) one person gets all knives, one person gets all forks, and one person gets all spoons. 更新: cccmartin46 part a 答案是對的，但part b不對。 我有答案但唔曉做！ thx~ 更新 2: part b 答案是6.93*10^ -4
最佳解答:
a) P(each person get one of each utensil) =(4!4!4!/12!)*6^4 =0.037402597 Remark: *6 because (f,s,k),(f,k,s),(s,f,k),(s,k,f),(k,s,f)or(k,f,s) also=1set utensil b) P(1 person gets all knives, 1 person gets all forks, and 1 person gets all spoons) =(4/12*3/11*2/10*4/9*3/8*2/7*4/6*3/5*2*4)*4! =(4!4!4!3!/12!)*4! =0.004155844156 Remark:*4! because (fff,sss,kkk,fsk)can be formed 24 diffenece sequences
其他解答:
a) P(each gets one of each utensil ) = 12/12 [1st person can get anything first] x 8/11 [There r 8 remaining pieces of cultery different from the first piece] x 4/10 [There r 4 remaining pieces of cultery different from both the first two pieces] x 9/9 [2nd person can get anything first] x 6/8 [There r 6 remaining pieces of cultery different from the previous piece] x 3/7 [There r 3 remaining pieces of cultery different from the previous two pieces] x 6/6 [3rd person can get anything first] x 4/5 [There r 4 remaining pieces of cultery different from the previous piece] x 2/4 [There r 2 remaining pieces of cultery different from the previous two pieces] [and the last person can get the remaining 3 different pieces in any order] = 1 * 8/11 * 4/10 * 1 * 6/8 * 3/7 * 1 * 4/5 * 2/4 = 72/4235 (or 0.017) b) P(one gets all knives, one gets all forks, and one gets all spoons) = 12/12 [1st person can get anything] x 3/11 [Only 3 same pieces of cultery remaining] x 2/10 [Only 2 same pieces remaining] x 8/9 [8 pieces out of 9 are different from what the 1st person got] x 3/8 [Only 3 same pieces of cultery remaining] x 2/7 [Only 2 same pieces remaining] x 4/6 [4 pieces out of 6 are different from what the 1st and 2nd persons got] x 3/5 [Only 3 same pieces of cultery remaining] x 2/4 [Only 2 same pieces remaining] [and the last person can get the remaining 3 different pieces in any order] = 1 * 3/11 * 2/10 * 8/9 * 3/8 * 2/7 * 4/6 * 3/5 * 2/4 = 2/4235 (or 0.00047)
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