Probability of event appearing at least 2 times

Question

The average percentage of defective product is 10%. There were released 100 units of this product. What is a probability that there will be at least 2 defective products among them?

I tried to solve this with Bernoulli formula. So, lets say $m$ is a number of defective products. I need to find $P(m \geq 2)$. We can rewrite it like this: $P(m \geq 2) = 1 - P(m < 2) = 1 - P(m=0) - P(m=1)$. So, $P(m=0)= {100\choose 0} (0,9)^{100}$, $P(m=1)= {100\choose 1} (0,1)^{1} (0,9)^{99}$.

So, $P(m \geq 3) = 1 - {100\choose 0} (0,9)^{100} - {100\choose 1} (0,1)^{1} (0,9)^{99} = 1 - 0,00003 - 10 \cdot 0,00003 = 0,99967$

It seems like too high probability. Did i do everything right?

Answer 1

Your computation is correct, I get the same result. Also, it does not seem too high to me, since it is extremely unlikely that there is only one or even zero defective products among 100 at such a high deficiency rate.

Answer 2

You did everything right (except that you mistyped $10$ instead in $100$ for the evaluation of ${100\choose 1}$, but the final result is ok).

It's not an unexpected result. If the failure rate is $10\%$, then in $100$ samples we have on average $10$ failures . So, we can guess, it's very probable that we have at least $2$ failures.

More in detail, it's a Binomial distribution, the standard deviation is $\sqrt{n p (1-p)}=3$, and it resembles a Gaussian, hence we can expect that more than $99.7\%$ of the ocurrences happen in the range $[7,13]$