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?

Question

Jaylon Carrillo

Mathematics

24 April, 14:48

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?

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Answers (1)

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Alanna Brennan · 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.