Math probability combination explanation

Question

A group of four components is known to contain two defectives. An inspector tests the components one at a time until the two defectives are located. once she locates the two defectives, she stops testing, but the second defective is tested to ensure accuracy. let Y denote the number of the test on which the second defective is found. Find the probability distribution for Y.

I know the answers are P[2] = 2/4 1/3 = 1/6 P[3] = 2c1(2/4 *2/3) * 1/2 = 1/3 P[4] = 3c2*(2/4 * 1/3 * 2/2) *1/1 = 1/2

But could someone please explain where those fractions came from?

Answer 1

For $Y=2$: (both defectives are found in the first two inspections) $$P(Y=2)=\text{(prob. of first defective) } \cdot \text{(prob. of second defective) }=\frac{2}{4} \cdot \frac{1}{3}.$$ The reason is: the prob.e of finding the first defective is $2$ out of total of $4$ components. Now once that is found there is only one defective left, hence the prob. of finding the second one now is $1$ out of a total of $3$.

For $Y=3$. You need to break it into two cases: $GDD$ and $DGD$. Where $GDD$ means the first component tested was "G"ood but the second was "D"efective and the third was "D"efective as well. You can guess what $DGD$ means. Observe that there cannot be $DDG$ because then the test would have ended with two trials only.

So $$P(Y=3)=\underbrace{\binom{2}{1}}_{\text{choosing how to place $G$}}\quad \underbrace{\left(\frac{2}{4}\cdot\frac{2}{3}\right)}_{\text{prob of $GD$}} \quad \cdot \quad \underbrace{\frac{1}{2}}_{\text{prob. of having the last component $D$}}.$$

I hope this will help.

NOTE:

For $Y=4$: you need to have the last tested component be $D$ and the first three should have 2$G's$ and 1$D$.

choosing a place for the first $D$: $\binom{3}{1}$, now fill in the details..

Answer 2

The numbers are small, so we do not need general theory to do the calculation. The random variable $Y$ can take on values $2$, $3$, or $4$.

$\Pr(Y=2)$: We need to have the first defective, and the second. The probability the first is defective is $\frac{2}{4}$. Given the first was defective, there are $3$ items left, of which $1$ is defective. So the probability the second is defective given the first was is $\frac{1}{3}$, and therefore $\Pr(Y=2)=\frac{2}{4}\cdot\frac{1}{3}$.

$\Pr(Y=3)$: This can happen in $2$ ways: (i) BGB (bad, good, bad) or GBB. By the same reasoning as above, the probability of BG is $\frac{2}{4}\cdot\frac{2}{3}$. If we started with BG, the probability the next is B is $\frac{1}{2}$. So the probability of BGB is $\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$. Similar reasoning shows the probability of GBB is $\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$, the same. So we need to multiply $\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}$ by $2$.

$\Pr(Y=4)$: This happens with the patterns GGBB, GBGB, and BGGB. It turns out that each has probability $\frac{2}{4}\cdot \frac{1}{3}$, giving probability $3\cdot\frac{2}{4}\cdot\frac{1}{3}$.

Remark: The probabilities must add up to $1$, so when we have found two of them, the other can be found by arithmetic. In this case, the probabilities simplify to $\frac{1}{6}$, $\frac{1}{3}$, and $\frac{1}{2}$.

There is a nice formula for the general case, where we have $b$ bad and $g$ good, and want to find the probability that the $m$-th bad is found on the $n$-th trial. However, since we only had three cases, direct calculation is easier.