機率算數題~求解

(1)
P(不良品) = 10% = 0.1
P(非不良品) = 1 - 0.1 = 0.9

P(8樣品中2個不良品)
= 8C2 x (0.1)² x(0.9)⁶
= 0.1488 (四位有效數字)


(2)
P(8樣品中至少2個不良品)
= 1 - P(8樣品中0個不良品) -P(8樣品1個不良品)
= 1 - [8C0 x (0.1)⁰ x(0.9)⁸] - [8C1 x (0.1)¹ x (0.9)⁷]
= 0.1869 (四位有效數字)


(3)
P(8樣品中不多於2個不良品)
= P(8樣品中0個不良品) +P(8樣品1個不良品) +P(8樣品中2個不良品)
= [1 - P(8樣品中至少2個不良品)]+ P(8樣品中2個不良品)
= (1 - 0.1869) + 0.1488
= 0.9619 (四位有效數字)


(4)
P(8樣品中少於2個不良品)
= 1 - P(8樣品中至少2個不良品)
= 1 - 0.1869
= 0.8131

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到下面的網址看看吧

▶▶http://candy5660601.pixnet.net/blog