Question

A special tool manufacturer has 50 customer orders to fulfill. Each order requires one special part that is purchased from a supplier. However, typically there are 2% defective parts. The components can be assumed to be independent. If the manufacturer stocks 52 parts, what is the probability that all orders can be filled without reordering parts

Answer 1

Answer:

0.65463

Step-by-step explanation:

From the given question:

It is stated that 2% of the parts are defective (D) out of 50 parts

Therefore the probability of the defectives;

i.e  p(defectives) = $\dfrac{N(D)}{N(S)}$

p(defectives) = $\dfrac{2}{50}$

p(defectives) = 0.04

The probability of the failure is the P(Non-defectives)

p(Non-defectives) =  1 - P(defectives)

p(Non-defectives) =  1 - 0.04

p(Non-defectives) =  0.96

Also , Let Y be the number of non -defective out of the 52 stock parts.

and we need Y ≥ 50

P( Y ≥ 50) , n = 52 , p = 0.96

P( Y ≥ 50) = P(50 ≤ Y ≤ 52) = P(Y = 50, 51, 52)

= P(Y = 50) + P(Y =51) + P(Y=52)    (disjoint events)

P(Y = 50) = $(^{52}_{50}) ( 0.96)^{50}(1-0.96)^2$

$P(Y = 50) = 1326 (0.96)^{50}(0.04)^2$

P(Y = 50) = 0.27557

P(Y = 51) =$(^{52}_{51}) ( 0.96)^{51}(1-0.96)^1$

$P(Y = 51) = 52(0.96)^{51}(0.04)^1$

P(Y = 51) = 0.25936

(Y = 52) =$(^{52}_{52}) ( 0.96)^{52}(1-0.96)^0$

$P(Y = 52) = 1*(0.96)^{52}(0.04)^0$

P(Y = 52) = 0.1197

∴

P(Y = 50) + P(Y =51) + P(Y=52) = 0.27557 + 0.25936 + 0.1197

P(Y = 50) + P(Y =51) + P(Y=52) = 0.65463