機率 PROBABILITY

2007-10-24 9:16 am

at a grocery store, eggs come in cartons that hold a dozen eggs. Experience indicates that 78.5% of the cartons have no broken eggs, 19.2% have one broken egg , 2.2%have two broken eggs , 0.1% have three broken eggs, and that the percentage of cartons with four or more broken eggs is negligible. An egg selected at random from a carton is found to be broken. What is the probability that this egg is the only broken one in the carton?

更新1:

答案是 0.803

回答 (3)

heatthespurs

2007-10-25 12:25 am

✔ 最佳答案

Let X be the no of broken egg in the carton

The question ask us to find P(X=1 | one egg selected at random from the carton is broken), where

P(X=1 | one egg selected at random from the carton is broken)

= P(X=1 AND “one egg selected at random from the carton is broken”) / P(“one egg selected at random from the carton is broken”) …… (*)

P(X=0 AND “one egg selected at random from the carton is broken”) = 78.5% * (0/12)
P(X=1 AND “one egg selected at random from the carton is broken”) = 19.2% * (1/12)
P(X=2 AND “one egg selected at random from the carton is broken”) = 2.2% * (2/12)
P(X=3 AND “one egg selected at random from the carton is broken”) = 0.1% * (3/12)

Therefore,
P(“one egg selected at random from the carton is broken”)
= 78.5% * (0/12) + 19.2% * (1/12) + 2.2% * (2/12) + 0.1% * (3/12)
= 0.0199167

Therefore, put the value into (*)

P(X=1 | one egg selected at random from the carton is broken)

= P(X=1 AND “one egg selected at random from the carton is broken”) / P(“one egg selected at random from the carton is broken”)

= 19.2% * (1/12) / 0.0199167

= 0.803346

參考: Standard conditional probability question, ***must know question*** for exam ^_^

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Prince

2007-10-25 1:22 am

0.803

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kevin

2007-10-24 9:27 am

19.2% / ( 19.2%+2.2%+0.1%)
= 19.2/21.5
= 89.3%

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