Question

From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find the Mean of X.

Answer 1

Find the probability of drawing x number of defective items in the sample. Since there are 2 defective items in the sample, x can vary from 0 to 2. This generates the probability mass function (or PMF) of the random variable X. Thus, if f(x) is the PMF of X, the mean or expectation of the random variable X is $\sum\limits_{\text{x = 0}}^{2}{\text{x}\cdot \text{f}\left( \text{x} \right)}$.
X is the number of defective items in the sample drawn of size 4 from the lot of 6 items containing 2 defective items. Thus, X takes the values x = 0, 1, 2.
Define, P(X = x) = Probability of drawing x defective items in the sample.
= $\dfrac{\text{No}\text{. of cases where x defective items are drawn}}{\text{Total no of ways a sample can be drawn}}$

Complete step-by-step answer:
Since, 4 items are drawn out of 6 items in the lot, total number of ways in which a sample can be drawn${}_{}^6C_4^{}$
Now, as x defective items are drawn in a sample of size 4, no of non-defective items = (4 – x)
Therefore, number of cases where x defective items are drawn ${}_{}^2C_x^{} \times {}_{}^4C_{4 - x}^{}$
${\text{P}}\left( {{\text{X}} = {\text{x}}} \right) = \dfrac{{{}_{}^2{\text{C}}_{\text{x}}^{} \times {}_{}^4{\text{C}}_{4 - {\text{x}}}^{}}}{{{}_{}^6{\text{C}}_4^{}}},\;{\text{x}} = 0,1,2$
This is the probability mass function of X.
$\therefore$ Mean (or Expectation) of X
$\begin{aligned} &\mathop { = \sum }\limits_{x = 0}^2 x.P\left( {X = x} \right) \\\ &\mathop { = \sum }\limits_{x = 1}^2 x.P\left( {X = x} \right)\;\;\left( {0.P\left( {X = 0} \right) = 0} \right) \\\ &= 1.\left( {\dfrac{{{}_{}^2C_1^{} \times {}_{}^4C_3^{}}}{{{}_{}^6C_4^{}}}} \right) + 2.\left( {\dfrac{{{}_{}^2C_2^{} \times {}_{}^4C_2^{}}}{{{}_{}^6C_4^{}}}} \right) \\\ &= \dfrac{{1 \times 2 \times 4 + 2 \times 1 \times 6}}{{\dfrac{{6 \times 5}}{{2 \times 1}}}} \\\ &= \dfrac{{20}}{{15}} = \dfrac{4}{3} \\\ \end{aligned}$
Thus, the mean of the random variable X is $\dfrac{4}{3}$.

Note: In this problem, from the PMF of the random variable X, it is clear that X follows a Hypergeometric distribution with parameters N = 6, n = 4 and p $=\text{ }\dfrac{2}{6}\text{ = }\dfrac{1}{3}$. The mean of a hypergeometric distribution with parameters N, n and p is np. Here also, $\text{np = 4 x }\dfrac{1}{3}\text{ = }\dfrac{4}{3}$. Thus, we can cross examine that our answer is correct.