Question

Suppose that 2% of produced computers by a company have defects, and defects occur independently of each other. a) Find the probability of exactly 2 defective computers in a shipment of 100 computers. b) The probability that a computer support specialist needs to test at least N computers in order to find one defective computer is 0.6676. Find N.

Answer 1

Answer:

a)  P(exactly two defectives) = 0.2734

b) N $\simeq$ 55

Step-by-step explanation:

From the given information:

P(exactly two defectives) = P(X =2) = $^{100}C_2 \times ( 0.02)^2 \times(1-0.02)^{100-2}$

$= \dfrac{100!}{2!(100-2)!} \times ( 0.02)^2 \times(0.98)^{98}$

$= 4950 \times 4 \times 10^{-4} \times 0.1380878341$

= 0.2734

Thus, P(exactly two defectives) = 0.2734

b)

To find:

P(X ≥ 1 )

let X be the random variable that obeys a binomial distribution, X represents the number of defectives,

∴

$X \sim B ( N, 0.02)$

$P(X \ge 1 ) = 1 - P(X < 1) \\ \\ P(X \ge 1 ) = 1 - P(X = 0)$

$0.6676 = 1 - ^NC_0 \times (0.02)^0 (0.98)^N$

$0.6676 = 1 - (0.98)^N$

$(0.98)^N = 1 - 0.6676$

$(0.98)^N = 0.3324$

N log (0.98) = log (0.3324)

$N = \dfrac{log \ 0.3324}{log \ 0.98}$

N = 54.51824841

N $\simeq$ 55