Question

The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 10%. If 14 calculators are selected at random, what is the probability that less than 5 of the calculators will be defective?

Answer 1

Solution

Given that ,

p = 0.10

1 - p = 0.90

n = 14

x < 5

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X < 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)

= ((14! / 0! (14-0)!) * 0.10⁰ * (0.90)^14-0 + ((14! / 1! (14-1)!) * 0.10¹ * (0.90)^14-1+ ((14! / 2! (14-2)!) * 0.10² * (0.90)^14-2 + ((14! / 3! (14-3)!) * 0.10³ * (0.90)^14-3+ ((14! / 4! (14-4)!) * 0.10⁴ * (0.90)^14-4  

= 0.2288 + 0.3559 + 0.257 + 0.1142 + 0.0349

= 0.9908

Probability = 0.9908

The probability that less than 5 of the calculators will be defective is 0.9908