What is expected value of having damaged PC if we choose 4 out of 8, where 3 of them is damaged.

Question

I don't get the logic for this again. So the probability that the PC is damaged is $\frac{3}{8}$ and four random PC are being selected out of these eight. Then the expect value of damaged PC of all 4 selected would be $\frac{3}{8} \cdot 4 = \frac{3}{2}=1.5$. I am not sure if it's correct. I also need to write a mathematical dispersion for this problem and am not sure how to do it.

Answer 1

The term "dispersion" in relation to the statistics of a population is somewhat vague, since there are several well-defined statistical quantities that may be taken as measures of it. Assuming that what you're being asked for is the standard deviation, $\ \sigma\ $, or equivalently, its square, $\ \sigma^2\ $, the variance, you can calculate it from the formula \begin{align} \sigma^2 &= \mathbb{E}\left(D^2\right)-\mathbb{E}\left(D\right)^2\\ &=\sum_{d=1}^3 d^2\mathbb{P}(D=d)-1.5^2\ , \end{align} where $\ D\ $ is the number of damaged PCs in your sample of $4$.

Unfortunately, the observations made in JMoravitz's comments, which enable you to simplify the calculation of $\ \mathbb{E}\left(D\right)\ $, don't help much in the calculation of $\ \mathbb{E}\left(D^2\right)\ $. If you want to calculate the latter quantity, it doesn't seem possible to avoid having to calculate the probabilities $\ \mathbb{P}(D=d)\ $. There are $\ 3\choose\ d\ $ subsets of size $\ d\ $ of the set of $3$ damaged PCs, and $\ 5\choose\ 4-d\ $ subsets of size $\ 4-d\ $ of the $5$ undamaged PCs. Thus, there are $\ {3\choose\,d}{5\choose\ 4-d}\ $ ways of choosing $4$ PCs, exactly $\ d\ $ of which are damaged, from the original set of $8$. The total number of ways of choosing $4$ PCs from the $8$, which you are apparently supposed to assume are equally likely to have occurred, is $\ 8\choose\ \ 4\ $. Therefore, $\ \mathbb{P}(D=d)=\frac{{3\choose\,d}{5\choose\ 4-d}}{8\choose\ 4}\ $, and \begin{align}\sigma^2&=\frac{1^2 {3\choose1}{5\choose 3}+2^2 {3\choose2}{5\choose 2}+3^2 {3\choose 3}{5\choose 1}}{8\choose 4}-1.5^2\\ &=\frac{69}{14}-2.25\\ &\approx 2.68\ , \end{align} and the standard deviation, $\ \sigma\approx1.64\ $