Question

The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 15%. If 16 calculators are selected at random, what is the probability that more than 5 of the calculators will be defective?

Answer 1

Here, n = 16, p = 0.15, (1 - p) = 0.85 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X > 5).
P(X <= 5) = (16C0 * 0.15^0 * 0.85^16) + (16C1 * 0.15^1 * 0.85^15) + (16C2 * 0.15^2 * 0.85^14) + (16C3 * 0.15^3 * 0.85^13) + (16C4 * 0.15^4 * 0.85^12) + (16C5 * 0.15^5 * 0.85^11)
P(X <= 5) = 0.0743 + 0.2097 + 0.2775 + 0.2285 + 0.1311 + 0.0555
P(X <= 5) = 0.9766

P(x >5) = 1- P(x < =5)
= 1- 0.9766
= 0.0234