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.
probaaility
2007-06-07 12:16 pm
回答 (2)
2007-06-07 5:39 pm
✔ 最佳答案
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
2007-06-08 1:41 am
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)
= 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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