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Ten items are to be sampled from a lot of 60. If more than one is defective, the lot will be rejected. Find the probability that the lot will be rejected in each of the following cases. a) The number of defective items in the lot is 5. b) The number of defective items in the lot is 10. c) The number of defective items in the lot is 20.

Respuesta :

Answer:

(a) The probability of that the lot will be rejected is 0.1904.

(b)The probability of that the lot will be rejected is 0.5315 .

(c)The probability of that the lot will be rejected is 0.7026.

Step-by-step explanation:

The formula of Hypergeometric distribution is

[tex]P(X=x)=\frac{(^{R}C_x)( ^{N-R}C_{n-x})}{^NC_n}[/tex]

N= number of population

R= The number of success event

n= sample of size

(a)

Given that, N=  60, R=5 and n=10

More than 1 is defected, the lot will be rejected.

X is random variable and defined the number of rejected item.

P(X>1)=1-P(X≤1)

        =1 - P(X=0) - P(X=1)

        [tex]=1- \frac{(^5C_0)(^{60-5}C_{10-0})}{^{60}C_{10}}-\frac{(^5C_1)(^{60-5}C_{10-1})}{^{60}C_{10}}[/tex]

        =1 - 0.3879 - 0.4217

          =0.1904

The probability of that the lot will be rejected is 0.1904.

(b)

N=  60, R=10 and n=10

X is random variable and defined the number of rejected item.

P(X>1)=1-P(X≤1)

        =1 - P(X=0) - P(X=1)

        [tex]=1- \frac{(^{10}C_0)(^{60-10}C_{10-0})}{^{60}C_{10}}-\frac{(^{10}C_1)(^{60-10}C_{10-1})}{^{60}C_{10}}[/tex]

       =1-0.1362-0.3323

      =0.5315

The probability of that the lot will be rejected is 0.5315 .

(c)

N=  60, R=20 and n=10

X is random variable and defined the number of rejected item.

P(X>1)=1-P(X≤1)

        =1 - P(X=0) - P(X=1)

        [tex]=1- \frac{(^{20}C_0)(^{60-20}C_{10-0})}{^{60}C_{10}}-\frac{(^{20}C_1)(^{60-20}C_{10-1})}{^{60}C_{10}}[/tex]

       =1 -0.2249 - 0.0725

       =0.7026

The probability of that the lot will be rejected is 0.7026.

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