Question

The J.O. Supplies Company buys calculators from a Korean supplier. The probability of a defective calculator is 30%. If 14 calculators are selected at random, what is the probability that more than 6 of the calculators will be defective?

Answer 1

Let p , probability of success that calculator will be defective is 0.30.

Given : n = 14, p = 0.30

Using Binomial Distribution,

P(x = r) = pr (1 - p)n - r

We need to find ,

P( x > 6) = 1 - P(x 6)

P(x 6) = P( x = 0) + P(x = 1) + .....+ P(x = 6)

= + + + + + +

= + + + + + +

= +   + + + + +

= 0.00678 + 0.04069 + 0.11336 + 0.19433 + 0.22903 + 0.19631 + 0.12620

P( x 6) = 0.9067

P(x > 6) = 1 - 0.9067

= 0.0933

Thus , there is 0.0933 probability that atleast 6 calculators will be defective out of 14.

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