Question

Warmliquid corp., a manufacturer of water heaters, produces 30 water heaters per week. past records indicate that 10% of total water heaters produced in a week are likely to be defective. suppose a quality check was conducted on a sample of six water heaters. what is the probability that four out of six water heaters will be defective?​

Answer 1

Let us use the Binomial Probability:

P = nCr * p^r * q^(n – r)

where,

n = total samples = 6

r = defective samples = 4

p = chance of defective = 10% = 0.10

q = 1 – p = 0.90

Solving for P:

P = 6C4 * 0.10^4 * 0.90^2

P = 1.215 x 10^-3 = 0.1215%

Only 1.215 x 10^-3 or 0.1215% chance that 4 out of 6 will be defective.