Name: Binomial Probability Distribution: Mean and Variance
Uploaded: 2019-07-19
Duration: 6 min 31 s
Description: A binomial distribution can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple time...

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: Variance of

X

.

Accepted Answer

Given - Defective items = $2$ ,

Hence, Non-defective items = $4$

Therefore, Probability of defective items = $X$ and non-defective items respectively is $\frac{4}{6}$

Therefore, $r$ items can be drawn from $n$ items in ${}^{n}C_{r}$ ways.

Since our random variable represents value of defective items being drawn,

Since $X = 0$ (no defective items drawn)

$\Rightarrow P (X = 0) = \frac{{}^{4}C_{4} \times {}^{2}C_{0}}{{}^{6}C_{4}} = \frac{1}{15}$

And $X = 1$ (one defective item drawn)

$\Rightarrow P (X = 1) = \frac{{}^{2}C_{1} {}^{4}C_{3}}{{}^{6}C_{4}} = \frac{8}{15}$

And lastly $X = 2$ (two defective items drawn)

$6$

Hence, our probability distribution will be as follows -

$X$ $0$ $1$ $2$

$P (X)$ $\frac{1}{15}$ $\frac{8}{15}$ $\frac{6}{15}$

And, we know that Mean of $X$ is given by
$E (X) = \sum_{i = 0}^{2} x_{i} p_{i}$
$= x_{0} p_{0} + x_{1} p_{1} + x_{2} p_{2}$
$= (0) (\frac{1}{15}) + (1) (\frac{8}{15}) + 2 \times \frac{6}{15}$
$= \frac{20}{15}$
Hence, Mean of $X$ $= \frac{4}{3}$ .

Now,

Variance of $X$ $= \sum_{i = 1}^{2} x_{i}^{2} p (x_{i}) - {(\sum_{i = 1}^{2} x_{i} p (x_{i}))}^{2}$

$= {(0)}^{2} \times \frac{1}{15} + 1 \times \frac{8}{15} + 4 \times \frac{6}{15} - {(\frac{4}{3})}^{2}$

$= \frac{8}{15} + \frac{24}{15} - \frac{16}{9}$

$= \frac{32}{15} - \frac{16}{9}$

$= \frac{16}{45}$

Hence, Variance of $X = \frac{16}{45}$

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: Variance of $X$ .

Important Questions on Probability

Learn and Prepare for any exam you want !

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: Variance of X.

Important Questions on Probability

Learn and Prepare for any exam you want !

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: Variance of $X$ .