Question

Among 12 metal parts produced in a machine shop, 3 are defective. If a random sample of 5 of these metal parts is selected, find:

(a) The probability that this sample will contain at least two defectives. (Do not round the intermediate answers. Round your answer to 4 decimal places.)

(b) The probability that this sample will contain at most one defective. (Round your answer to 4 decimal places.)

Answer 1

Given that among 12 metal parts produced in a machine shop, 3 are defective. A random sample of 5 of these metal parts is selected.

Now, 3 of 12 metal parts are defective. Thus, the probability of a defective part, p = 3/12 = 0.25

Now, a random sample of 5 parts are selected. Thus, n = 5.

From binomial distribution, we have,

, for x = 0,1,2,3,...

a) Here, we have to find the probability that this sample will contain at least 2 defectives. Thus, we have to find P(X2). Thus, we have to find P(X=2)+P(X=3).

Thus, we have n = 5, p = 0.25

Now, P(X = 2) = = = [(5!)/{2!(5-2)!}]*0.026367 = [(5*4*3!)/(2!*3!)]*0.026367 = 10*0.026367 = 0.26367.

Thus, P(X = 2) = 0.26367

Now, = [(5*4)/(2*1)]*0.008789 = 10*0.008789 = 0.08789

Thus, P(X = 3) = 0.08789.

Thus, P(X = 2) + P(X = 3) = 0.26367+0.08789 = 0.35156 = 0.3516(rounded up to four decimal places).

Thus, P(X2) = 0.3516 .

Thus, the probability that this sample will contain at least two defectives = 0.3516 .

b) Here, we have to find the probability that this sample will contain at most one defective. Thus, we have to find P(X1). Thus, we have to find P(X=0)+P(X=1).

Now, = 1*1*0.2373 = 0.2373 .

Thus, P(X = 0) = 0.2373.

Now, = 5*0.25*0.3164 = 0.3955.

Thus, P(X = 1) = 0.3955 .

Thus, P(X = 0) + P(X = 1) = 0.2373 + 0.3955 = 0.6328.

Thus, P(X1) = 0.6328 .

Thus, the probability that this sample will contain at most one defective = 0.6328 .