Question

Gears produced by a grinding process are categorized either as conforming (suitable for their intended purpose), degraded (unsuitable for the intended purpose but usable for another purpose), or scrap (not usable). Suppose that 75% of the gears produced are conforming, 11% are degraded, and 14% are scrap. Ten gears are selected at random.

What is the probability that eight or more are not scrap?

What is the probability that more than two are either degraded or scrap?

What is the probability that exactly nine are either conforming or degraded?

Answer 1

P(C) = 0.75
P(D) = 0.11
P(S) = 0.14

a)
Here, n = 10, p = 1 - 0.14 = 0.86, (1 - p) = 0.14 and x = 8
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 8).
P(X >= 8) = (10C8 * 0.86^8 * 0.14^2) + (10C9 * 0.86^9 * 0.14^1) + (10C10 * 0.86^10 * 0.14^0)
P(X >= 8) = 0.2639 + 0.3603 + 0.2213
P(X >= 8) = 0.8455

b)
Here, n = 10, p = 0.25, (1 - p) = 0.75 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 1).
P(X <= 1) = (10C0 * 0.25^0 * 0.75^10) + (10C1 * 0.25^1 * 0.75^9)
P(X <= 1) = 0.0563 + 0.1877
P(X <= 1) = 0.244

required probability = 1 - 0.244 = 0.7560

c)
Here, n = 10, p = 0.86, (1 - p) = 0.14 and x = 9
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 9)
P(X = 9) = 10C9 * 0.86^9 * 0.14^1
P(X = 9) = 0.3603