請問各位大大,底下題目條件機率的算法,謝謝

2015-02-13 7:58 pm
一個盒子中有998個黑球及2個白球,若自盒中隨機挑選500球,令x為其中白球之數量,試求條件機率P(x=1 | x=1或x=2)?

回答 (3)

2015-02-13 9:24 pm
✔ 最佳答案
P( x = 1 | x = 1 或 x = 2 )

= P[ x = 1 及 (x = 1 或 x = 2) ] / P(x = 1 或 x = 2)

= P( x = 1 ) / P(x = 1 或 x = 2)

= [(₂C₁)(₉₉₈C₄₉₉)/(₁₀₀₀C₅₀₀)] / [(₂C₁)(₉₉₈C₄₉₉)/(₁₀₀₀C₅₀₀) + (₂C₂)(₉₉₈C₄₉₈)/(₁₀₀₀C₅₀₀)]

= [(₂C₁)(₉₉₈C₄₉₉)] / [(₂C₁)(₉₉₈C₄₉₉) + (₂C₂)(₉₉₈C₄₉₈)]

= 0.667111408


2015-02-13 14:55:36 補充:
謝謝兩位的意見!

請讓我加入補充並嗚謝~~!!!

﹝。◕‿◕。◕‿◠。﹞

2015-02-13 22:55:53 補充:
謝謝 阿番 網友指出:

答案以分數表示為1000/1499

2015-02-13 22:56:57 補充:
謝謝 ☂雨後晴空☀ 網友提出另解:

P(x=1 | x=1或x=2)

    P(x=1)
= ───────────
  P(x=1) + P(x=2)

        (500C1) (500C1) / 1000C2
= ─────────────────────────────────────
  (500C1) (500C1) / 1000C2 + (500C0) (500C2) / 1000C2

2015-02-13 22:57:17 補充:
     (500) (500)
= ────────────────────
  (500) (500) + (1) (500×499/2)


    500
= ──────────
  500 + 499/2


  1000
= ────
  1499
2015-02-13 9:37 pm
P(x=1 | x=1或x=2)

    P(x=1)
= ───────────
  P(x=1) + P(x=2)

        (500C1) (500C1) / 1000C2
= ─────────────────────────────────────
  (500C1) (500C1) / 1000C2 + (500C0) (500C2) / 1000C2

2015-02-13 13:37:34 補充:
     (500) (500)
= ────────────────────
  (500) (500) + (1) (500×499/2)


    500
= ──────────
  500 + 499/2


  1000
= ────
  1499
2015-02-13 9:33 pm
答案以分數表示為1000/1499


收錄日期: 2021-04-21 22:29:31
原文連結 [永久失效]:
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