Question25 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

Rogelio Vega

Mathematics

24 December, 11:10

Question25 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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Adeline Johnson · Answer 1

To solve this problem, we use the binomial probability equation.

P = [n! / (n - r) ! r!] p^r q^ (n - r)

where,

n is the total number of samples = 6

r is the number of defective = 4

p is the probability it will be defective = 0.10

q is the probability it wont be = 0.90

P = [6! / (6 - 4) ! 4!] (0.10) ^4 * (0.90) ^ (6 - 4)

P = 1.215 x 10^-3 = 0.1215%