Determine the probability that at least 4 brake pads are defective in an order of 200 pads.

Question

In a manufacturing process, 0.2% of the brake pads that come off of the production line are defective. Determine the probability that at least 4 brake pads are defective in an order of 200 pads.

I’m very confused with this homework question. A solution to this question using binomial distributions with an explanation would be very helpful. Thank you!

Answer 1

We can generate the answer using the binomial formula or by using software where we enter n(the number of trials), x(the number of successes), and p(the probability). We also indicate whether this is cumulative (all values up to x), or just the value at x.

The binomial formula is C(n,x)p^x(1-p)^n-x where C(n,x) = n!/(x!(n-x)!)

In this problem, n = 200 and p = .002

Then, the probability of at least 4 is 1 minus the probability of at most 3, or 1 - (P(0)+P(1)+P(2)+P(3))

Thus, using the bin. formula, 1 - (C(200,0)p⁰(1-p)²⁰⁰ + C(200,1)p¹(1-p)¹⁹⁹ + C(200,2)p²(1-p)¹⁹⁸ + C(200,3)p³(1-p)¹⁹⁷) =

1-(1*1*.6701 + 200*.002*.6714 + 19900*.000004*.6727 + 1313400*.000000008*.6741) =

1-.6701-.2686-.0536-.0071 =

0.0007578 (showing additional significant digits here)

However, as I mentioned, with packages, you don't have to do these calculations.

With Excel, for example, we would simply enter

1 - binom.dist(3,200,.002,1) = 0.0007578

Note: if you calculate np, this is the expected value. In this case, np = 200 * .002 = .4.

We know from this that we will usually have 0 or 1 defective brakes.