Question

6.24 An electronic production line yields 95% satisfactory parts. Assume that the quality of each part is independent of the others. Four parts are tested. Determine the probability that (a) all parts tested are satisfactory. (b) at least two parts are satisfactory.

Answer 1

Answer: (a)0.8145

(b) 0.99951875

Step-by-step explanation:

Binomial probability formula to find the probability of getting success in x trials :-

$P(X)=^nC_xp^x(1-p)^{n-x}$ , where n is the total number of trials and p is the probability of success in each trial.

Given : The probability that the electronic production line yields satisfactory parts =0.95

The number of parts tested = 4

(a) The probability that all parts tested are satisfactory:-

$P(4)=^4C_4(0.95)^4(0.05)^{0}=(0.95)^4=0.81450625\approx0.8145$

(b) The probability that at least two parts are satisfactory :-

$P(x\geq2)=1-P(x\leq1)=1-(P(0)+P(1)))\\\\=1-(^4C_0(0.95)^0(0.05)^{4}+^4C_1(0.95)^1(0.05)^{3})\\\\=1-((0.05)^{4}+4(0.95)(0.05)^3)=0.99951875\approx0.9995$