F.4 Maths probability

The required probability is

　Pr( Box C | blue )

= Pr( Box C and blue ) / Pr( blue )

= Pr( Box C ) × Pr( blue | Box C ) /
　 { Pr( Box A ) × Pr( blue | Box A ) + Pr( Box B ) × Pr( blue | Box B ) + Pr( Box C ) × Pr( blue | Box C ) }

= (1/3) × (9/21) / [ (1/3) × (8/14) + (1/3) × (7/14) + (1/3) × (9/21) ]

= (9/21) / [ (8/14) + (7/14) + (9/21) ]

= (3/7) / [ (4/7) + (1/2) + (3/7) ]

= (6) / [ (8) + (7) + (6) ]

= 6/21

= 2/7


2014-05-27 18:14:10 補充：
Thanks for your input.

(｡◕‿◕｡)

2014-05-28 18:28:17 補充：
YTC：

The probability of A given B is Pr( A | B ) = Pr( A and B ) / Pr(B)

這個原則是要記熟的。

當中的原意是指：當已經知道 B 會發生，問 A 發生的機會。
所以 B 已發生，這個變成前提。
分母的部份也隨至有所變化。

2014-05-28 18:32:34 補充：
以下我舉一例解釋條件概率 (conditional probability) 的計算。

假設有以下十個數： { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
從中抽一個。

那麼 Pr( 3's multiple ) = #{3, 6, 9} / 10 = 3 / 10

如果現在問 Pr( 3's multiple | the number < 5 )
那麼可供選擇的數不是 10 個，而只是 4 個。
Prob = #{3} / #{1, 2, 3, 4} = 1 / 4

2014-05-28 18:33:11 補充：
驗證一下：
　Pr( 3's multiple | the number < 5 )
= Pr( 3's multiple and the number < 5 ) / Pr( the number < 5 )
= (1/10) / (4/10)
= 1/4

P([Box A] and [blue]) = (1/3) x [8/(6+8)] = 4/21
P([Box B] and [blue]) = (1/3) x [7/(7+7)] = 1/6
P([Box C] and [blue]) = (1/3) x [9/(9+12)] = 1/7

P(blue) = (4/21) + (1/6) + (1/7) = (8/42) + (7/42) + (6/42) = 1/2

P([box C] l [blue])
= P([box C] and [blue]) / P(blue)
= (1/7) / (1/2)
= 2/7