→ a016258:考慮 第一次抽中 第二次沒中 (1/5)*(3/4)*2 (一號或二號) 10/23 02:06
→ a016258:同理 第一次沒 第二次有 也是 所以總共就是(1/5)*(3/4)*4 10/23 02:07
→ nana0130:喔~~太感謝你了!^^ 10/23 02:23
Many manufacturers have quality control programs that include inspection
of incoming matericals for defects.
Suppose a computer manufacturer receives computer boards in lots of five.
Two boards are selected from each lot for inspection . We can represent
possible outcomes of the selection process by pairs. For example , the pair
(1,2) represent the selection of boards 1 and 2 for inspection.
b. Suppose the boards 1 and 2 are the only defective boards in a lot of five
Two boards are to be chosen at random. Define X to be the number of
defective boards observed among those inspected . Find the probability
distribution of X.
X=0 , p(0) = 3/5 * 2/4 = 0.3 (正確)
X=1 , p(1) = 1/5 * 1 (第一次就抽到一號，後面不管)
+ 4/5(第一次沒抽到一號)*1/4(第二次抽到一號) = 0.4
這個是錯的 ，解答是0.6
請問我哪邊算錯了勒????
謝謝你
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