Three percent of the computer chips produced by a machine are defective. Distribution

Question

Three percent of the computer chips produced by a machine are defective. These computer chips are delivered to retailers in packages of 20 chips each. Assume that the chips in any package as a random sample of 20 chips from the chips produced by this machine.

(a) A package is selected randomly. What is the probability that the package will contain at least 2 defective chips? Give your answer to four decimal places. Note: At least two means 2 or more.

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My attempt:

This passes most of the binomial experiment trials therefore Let $X$ be the number of defective chips. $X$ ~ $Binomial(n = 20, \theta=0.03)$

$P(X \ge 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1))$

$= 1 - \binom{20}{0}(0.03)^0(0.97)^{20} - \binom{20}{1}(0.03)^1(0.97)^{19} \approx 0.1198$

(b) We pick packages at random one by one and count how many defective chips are there in each package. If a package contains two or more defective chips, it is not considered as a good package. What is the probability that the tenth package selected is the third package containing at least two defective chips? Give your answer to four decimal places.

I don't get b, I'm sure we need the answer for (a) to do it, not sure how to go about it?

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Attempt:

Let $Y$ be the number of good packages before the third bad package. What experiment do I use? It fits the negative binomial definition but not sure how to use it

Answer 1

Given a random variable $Y\sim \text{NB}(r,p)$, then $P(Y=k)$ is the probability of having $k$ successes before the $r$-th failure occurs from a sequence of independent Bernoulli trials with parameter $p$.

In this case $p=P(X<2)\approx 1-0.1198$, as you found out in part (a). The problem asks for the third failure ($r=3$) to happen on the tenth try, so $k=7$.

Hence $$ P(Y=7)=\binom{k+r-1}{k}p^k(1-p)^r=\binom{9}{7}p^7(1-p)^3\approx 36(0.8802)^7(0.1198)^3=0.0253 $$

Answer 2

For convenience let $p$ denote the probabiliy that a package is not good.

So $p$ equals the probability that you calculated at a).

Then e.g. the probability that packages $2,6,10$ are not good while the others among the first $10$ are good is $p^3(1-p)^7$.

How many such possibilities exist? Well it comes to choosing $2$ packages among the first $9$. In the example we took $2$ and $6$.

So we end up with probability: $$\binom92p^3(1-p)^7$$