Probability

Tossing two fair coins once. Find the conditional probability that both

coins show tails given that

(a )the first coin shows a tail;

(b) at least one coin shows a tail

2 個解答

評分
  • 6 年前
    最愛解答

    Let Ti denote the event that the i-th coin shows a tail.

    (a)

    Pr(both coins show tails | the first coin shows a tail)

    = Pr(T1 and T2 | T1)

    = Pr(T1 and T2 and T1) / Pr(T1)

    = Pr(T1 and T2) / Pr(T1)

    = [(1/2)(1/2)] / (1/2)

    = 1/2

    (b)

    Pr(both coins show tails | at least one coin shows a tail)

    = Pr(T1 and T2 | T1 or T2)

    = Pr(T1 and T2 and "T1 or T2") / Pr(T1 or T2)

    = Pr( (T1 and T2 and T1) or (T1 and T2 and T2) ) / Pr(T1 or T2)

    = Pr( (T1 and T2) or (T1 and T2) ) / Pr(T1 or T2)

    = Pr(T1 and T2) / Pr(T1 or T2)

    = Pr(T1 and T2) / [ Pr(T1) + Pr(T2) - Pr(T1 and T2) ]

    = [(1/2)(1/2)] / [1/2 + 1/2 - (1/2)(1/2)]

    = (1/4) / (1 - 1/4)

    = 1/(4 - 1)

    = 1/3

    ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀ ✿ ❀

    Alternative Method

    There are only four possible situations, with equal chances:

    (1st T, 2nd T)

    (1st T, 2nd H)

    (1st H, 2nd T)

    (1st H, 2nd H)

    (a)

    Pr(both coins show tails | the first coin shows a tail)

    = Pr( {(1st T, 2nd T)} | {(1st T, 2nd T), (1st T, 2nd H)} )

    = Pr( {(1st T, 2nd T)} and {(1st T, 2nd T), (1st T, 2nd H)} ) /

    Pr({(1st T, 2nd T), (1st T, 2nd H)})

    = Pr( {(1st T, 2nd T)} ) / Pr({(1st T, 2nd T), (1st T, 2nd H)})

    = #{(1st T, 2nd T)} / #{(1st T, 2nd T), (1st T, 2nd H)}

    = 1/2

    (b)

    Pr(both coins show tails | at least one coin shows a tail)

    = Pr( {(1st T, 2nd T)} | {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} )

    = Pr( {(1st T, 2nd T)} and {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} ) / Pr( {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} )

    = Pr( {(1st T, 2nd T)} ) / Pr( {(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)} )

    = #{(1st T, 2nd T)} / #{(1st T, 2nd T), (1st T, 2nd H), (1st H, 2nd T)}

    = 1/3

    2015-02-01 13:00:35 補充:

    呢個問題問得好好!

    呢個係典型的 conditional probability 會造成的一些有趣的 paradox 的原因。

    〔稍後你有時間可以看看 Monte Hall Problem,你會見到很有趣!〕

    先答你的問題。

    我們明顯知道有四個情況:

    1. (T, T)

    2. (T, H)

    3. (H, T)

    4. (H, H)

    首先,已知 at least 1 tail,即是否決了 4.,但有可能是 1, 2, 3。

    在這三個選項中,問 both tail (即是 1) 的機就,就是三個之中的一個。

    而且機會均等,所以是 1/3。

    2015-02-01 13:10:22 補充:

    從數學看你詢問的那個分子:

    P(both tails AND at least 1 tail)

    = P( situation(T, T) AND situations{(T, H), (H, T), (T, T)} )

    = P( situation(T, T) )

    = (1/2)²

    = 1/4

    你想分子的時候,要用 AND 的概念去思考。

    因為你已經把 given 的意思轉化成公式:

    P( A | B ) = P( A and B ) / P(B)

    那麼你思考分子 P( A and B ) 的一刻就要想 AND,不要再想 given。

    2015-02-01 13:13:26 補充:

    為了令你更明白本題的精髓,我再舉一個常見的例子。

    假設男女機會均等。

    (a)

    在一個有兩個孩子的家庭中,已知年長的是男孩,問年幼的是男孩的機會。

    答案: 1/2

    (b)

    在一個有兩個孩子的家庭中,已知其中一個是男孩,問另一個是男孩的機會。

    答案: 1/3

    知否為何兩題的答案不一?

    這就是類似意見 002 的有趣之處。

    2015-02-01 13:14:55 補充:

    設符號標記為(年長,年幼)

    可能性:

    (男,男)

    (男,女)

    (女,男)

    (女,女)

    (a) 己知年長的是男孩,可能性是 { (男,男), (男,女) }

    (b) 已知其中一個是男孩,可能性是 { (男,男), (男,女), (女,男) }

    就是這個原因。

    2015-02-01 13:23:36 補充:

    以上我在 (a) 和 (b) 問的:

    {年幼的是男孩} 和 {另一個是男孩} 的機會,

    其實已經是在各題中相當於 {兩個皆是男孩} 的機會。

    看回原題:

    你詢問為什麼不是單單的乘 1/2 就是因為你以為情況是 (a) 的類型,但其實題目是問 (b) 的類型。

    {at least one coin shows a tail} 類似於 {其中一個是男孩}

    {both coins show tails} 類似於 {兩個皆是男孩}

    2015-02-02 22:47:58 補充:

    Cheers~

    ╭∧---∧╮

    │ .✪‿✪ │

    ╰/) ⋈ (\\╯

  • 6 年前

    P(both tails | at least 1 tail) = (1/2)^2 / (1-1/4)

    我想問點解 P(both tails AND at least 1 tail) 唔係 1/2 既?

    已知有一個tail, 咁要另外果個都係tail唔係只乘 1/2 就得喇咩?

    求指教, thanks!

    2015-02-02 20:55:58 補充:

    非常感謝你咁詳細既解釋!!!!!!!!!

    我下次會列晒啲結果出黎, 咁就一目了然~ ^^

    THX A MILLION!!

還有問題嗎?立即提問即可得到解答。