Question

please answer rounded to four decimal places as needed, thank you!

The probability that a part produced by a certain factory's assembly line will be defective is 0.012. Find the probabilities that in a run of 45 items, the following results are obtained. (a) Exactly 3 defective items (b) No defective items (c) At least 1 defective item

Answer 1

X : number of defective items,

X ~ bin (n,p)

where, n= 45

p = 0.012

q = (1-p) = (1-0.012) = 0.988

the pmf of the distribution be:-

$P(X=x)=\binom{45}{x}*0.012^x*0.988^{45-x}$

a).the probability of getting exactly 3 defective items is:-

$P(X=3)=\binom{45}{3}*0.012^3*0.988^{45-3}\approx\mathbf{0.0148}$

b).the probability of getting no defective items is:-

$P(X=0)=\binom{45}{0}*0.012^0*0.988^{45-0}\approx\mathbf{0.5808}$

c).the probability of getting at least one defective item is:-

$=P(X\geq 1)$

$=1-0.5808$ (from part b)

$=\mathbf{0.4192}$

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