F.5 M1 binomial distribution

2011-04-18 12:39 am

A certain type of electronic devices is produced by a factory with three production lines P, Q and R, which account for 40%, 20% and 40% of the total output of the factory respectively.

A random sample of 10 devices is drawn.

(a) Find the probability that, in the sample, the number of devices produced by P, Q and R are 3, 3 and 4 respectively.

Past records reveal that the proportion of defective devices produced by P, Q and R are respectively 15%, 20% and 10%, and the proportion of defective devices produced by the whole factory is 14%.

(b) If the number of devices produced by P, Q and R in the sample are 3, 3 and 4 respectively, find the probability that there are exactly two defective devices in the sample and only one of them was produced by P.

回答 (1)

myisland8132

2011-04-18 3:51 am

✔ 最佳答案

(a) Using multinomial distribution, the probability is

(10!)/(3!3!4!) * (0.4)^3(0.2)^3(0.4)^4 = 0.05505

(b) Required probability

= P(1 defective devices of P and 1 defective devices of Q ) + P(1 defective devices of P and 1 defective devices of R )

= C(3,1)(0.85)^2(0.15)C(3,1)(0.8)^2(0.2)(0.9)^4 + C(3,1)(0.85)^2(0.15)(0.8)^3C(4,1)(0.9)^3(0.1)

= 0.1305

Notice that the information"proportion of defective devices produced by the whole factory is 14%." is useless

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