Three boxes A, B and C each contains 12 rechargeable batteries. 1 of the batteries in Box A is fully charged, 2 of the batteries in Box B are fully charged and 3 of the batteries in Box C are fully charged.
Q. If a box is selected at random and then 2 batteries are selected at random from it, find the probability that at most one of the batteries selected is fully charged.
Ans. 97/99
Further Probability (F.5)
2013-02-05 9:20 am
回答 (2)
2013-02-05 10:16 am
✔ 最佳答案
F : fully chargedN : not fully charged
In Box A : 1F + 11N
In Box B : 2F + 10N
In Box C : 3F + 9N
P(0F 2N)
= P(Box A and NN) + P(Box B and NN) + P(Box C and NN)
= (1/3) x (11C2/12C2) + (1/3) x (10C2/12C2)+ (1/3)(9C2/12C2)
= (1/3) x [(11!/9!2!)/(12!/10!2!)] + (1/3)[(10!/8!2!)/(12!/10!2!)] + (1/3)[(9!/7!2!)/(12!/10!2!)]
= (10/36) + (30/132) + (6/33)
= (110/396) + (90/396) + (72/396)
= 272/396
= 68/99
P(1F 1N)
= P(Box A and 1F 1N) + P(Box B and 1F 1N) + P(Box C and 1F 1N)
= (1/3) x (11C1x1C1/12C2)+ (1/3) x (10C1x2C1/12C2)+ (1/3)(9C1x3C1/12C2)
= (1/3) x [11x1/(12!/10!2!)] + (1/3)[10x2/(12!/10!2!)] + (1/3)[9x3/(12!/10!2!)]
= (1/18) + (10/99) + (3/22)
= (11/198) + (20/198) + (27/198)
= 58/198
= 29/99
P(at most 1F)
= P(0F 2N) + P(1F 1N)
= (68/99) + (29/99)
= 97/99
參考: wanszeto
2013-02-06 5:41 pm
required probability
= (1 / 3) x [1] + (1 / 3) x [1 - (2C2 / 12C2)] + (1 / 3) x [1 - (3C2 / 12C2)]
= 97 / 99
= (1 / 3) x [1] + (1 / 3) x [1 - (2C2 / 12C2)] + (1 / 3) x [1 - (3C2 / 12C2)]
= 97 / 99
收錄日期: 2021-04-20 13:34:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130205000051KK00024