YTC asked in 科學及數學數學 · 6 years ago

F.4 Maths probability

One box is chosen ramdomly and a marble is drawn randomly from it.

Box A---no.of red marbles:6

no.of blue marbles:8

Box B---no.of red marbles:7

no.of blue marbles:7

Box C---no.of red marbles:12

no.of blue marbles:9

Given that the marble chosen is blue.

Find the probability that it is drawn from box C.

Ans. 2/7

Please help! I can only calculate 1/7......

Update:

RE: 知足常樂

不太明白如何由Pr( Box C | blue )轉到 Pr( Box C and blue ) / Pr( blue )

能否再解釋?

Thankyou!

2 Answers

Rating
  • 6 years ago
    Best Answer

    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

  • 土扁
    Lv 7
    6 years ago

    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

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