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# From a Lot of 15 Bulbs Which Include 5 Defective, a Sample of 4 Bulbs is Drawn One by One with Replacement. Find the Probability Distribution of Number of Defective Bulbs. - CBSE (Commerce) Class 12 - Mathematics

ConceptRandom Variables and Its Probability Distributions

#### Question

From a lot of 15 bulbs which include 5 defective, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.

#### Solution

Let getting a defective bulb in a trial be a success.
We have,

$p = \text{ probability of getting a defective bulb } = \frac{5}{15} = \frac{1}{3} \text{ and}$
$q = \text{ probability of getting non - defective bulb } = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$
$\text{ Let X denote the number of success in a sample of 4 trials . Then, }$
$\text{ X follows binomial distribution with parameters n = 4 and } p = \frac{1}{3}$
$\therefore P\left( X = r \right) =^{4}{}{C}_r p^r q^\left( 4 - r \right) = ^{4}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^\left( 4 - r \right) = \frac{^{4}{}{C}_r 2^\left( 4 - r \right)}{3^4}, \text{ where } r = 0, 1, 2, 3, 4$
$\text{ i . e .}$
$P\left( X = 0 \right) = \frac{^{4}{}{C}_0 2^4}{3^4} = \frac{16}{81},$
$P\left( X = 1 \right) = \frac{^{4}{}{C}_1 2^3}{3^4} = \frac{32}{81},$
$P\left( X = 2 \right) = \frac{^{4}{}{C}_2 2^2}{3^4} = \frac{24}{81},$
$P\left( X = 3 \right) = \frac{^{4}{}{C}_3 2^1}{3^4} = \frac{8}{81},$
$P\left( X = 4 \right) = \frac{^{4}{}{C}_4 2^0}{3^4} = \frac{1}{81}$

So, the probability distribution of X is given as follows:

 X: 0 1 2 3 4 P(X): $\frac{16}{81}$ $\frac{32}{81}$ $\frac{24}{81}$ $\frac{8}{81}$ $\frac{1}{81}$

Now,

$\text{ Mean } , E\left( X \right) = 0 \times \frac{16}{81} + 1 \times \frac{32}{81} + 2 \times \frac{24}{81} + 3 \times \frac{8}{81} + 4 \times \frac{1}{81}$
$= \frac{32 + 48 + 24 + 4}{81}$
$= \frac{108}{81}$
$= \frac{4}{3}$
$\text{ Note: We can also calculate the mean of the binomial distribution by }$
$\text{ Mean } , E\left( X \right) = \text{ np } = 4 \times \frac{1}{3} = \frac{4}{3}$

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Solution From a Lot of 15 Bulbs Which Include 5 Defective, a Sample of 4 Bulbs is Drawn One by One with Replacement. Find the Probability Distribution of Number of Defective Bulbs. Concept: Random Variables and Its Probability Distributions.
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