Suppose 2% of all cell phone chips are defective. we randomly select 100 of the 50,000 cell phone chips produced in a day. what is the chance that 0 defective chips will be found?

To solve this problem, we make use of the Binomial Probability equation. The formula is:

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

where,

n = the total number of samples tested = 100

r = the total number of defective chips = 0

p = probability of being defective = 0.02

q = probability of not being defective = 0.98

Hence, substituting the values:

P = [100! / 0! (100 – 0)!] 0.02^(0) * 0.98^(100)

P = [1] * 0.02^(0) * 0.98^(100)

P = 0.1326 = 13.26%

Therefore there is a 13.26% probability that 0 defective chips will be found.

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