simple probability problem

[匿名]

2013-02-05 11:22 am

a boy draws 5 pieces of calligraphy and 8 pieces of paintings, all of which are different. if he carelessly leaves three pieces of his art work on the bus, what is the probability that all of them are paintings, given that at least one of them is painting?

點解個答應係14/69? 列出步驟, THANKS

回答 (1)

用戶10012

2013-02-05 3:56 pm

✔ 最佳答案

By Law of Conditional probability,
P(All 3 are paintings | at least 1 is painting)
= P(All 3 are painting and at least 1 is painting)/P(At least 1 is painting)
P(All 3 are painting and at least 1 is painting) = P(All 3 are painting)
= (8/13)(7/12)(6/11).
P(At least 1 is painting) = P(1 is painting) + P(2 are painting) + P(All 3 are painting)
P(1 is painting) = 3 x (8/13)(5/12)(4/11). [Times 3 because it could be PCC, CPC or CCP].
P(2 are painting) = 3 x (8/13)(7/12)(5/11). [ Times 3 because it could be PPC, PCP or CPP].
So probability = (8 x 7 x 6)/[3 x 8 x 5 x 4 + 3 x 8 x 7 x 5 + 8 x 7 x 6]
= 42/(60 + 105 + 42) = 42/207 = 14/69.

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