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

Five bad oranges are accidentally mixed with $20$ good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.

Accepted Answer

We will find the probability distribution of the number of bad oranges drawn. Also, find the mean and variance of the distribution.

The total number of oranges $n (S) = 25$ .

bad oranges are $5$ .

Let $p =$ Probability of getting success(bad oranges) $=$ $\frac{5}{25} = \frac{1}{5}$

$∴ q = \frac{4}{5}$ , $n = 4$

Let the random variable $X$ is defined as the number of bad oranges drawn in a simple of four oranges drawn with replacement, so $X$ can takes values $0, 1, 2, 3, 4$ .

By using binomial distribution, we have,

$P (X = r) = {}^{n}C_{r} p^{r} q^{n - r}$

Now, probabilities are :

$P (X = 0) = {}^{4}C_{0} p^{0} q^{4 - 0} = {}^{4}C_{0} {(\frac{1}{5})}^{0} {(\frac{4}{5})}^{4 - 0} = {}^{4}C_{0} {(\frac{4}{5})}^{4} = \frac{256}{625}$

$P (X = 1) = {}^{4}C_{1} p^{1} q^{4 - 1} = {}^{4}C_{1} {(\frac{1}{5})}^{1} {(\frac{4}{5})}^{4 - 1} = {}^{4}C_{1} {(\frac{1}{5})}^{1} {(\frac{4}{5})}^{3} = \frac{256}{625}$

$P (X = 2) = {}^{4}C_{2} p^{2} q^{4 - 2} = {}^{4}C_{2} {(\frac{1}{5})}^{2} {(\frac{4}{5})}^{4 - 2} = {}^{4}C_{2} {(\frac{1}{5})}^{2} {(\frac{4}{5})}^{2} = \frac{6 \times 16}{625} = \frac{96}{625}$

$P (X = 3) = {}^{4}C_{3} p^{3} q^{4 - 3} = {}^{4}C_{3} {(\frac{1}{5})}^{3} {(\frac{4}{5})}^{4 - 3} = {}^{4}C_{3} {(\frac{1}{5})}^{3} {(\frac{4}{5})}^{1} = \frac{4 \times 4}{625} = \frac{16}{625}$

$P (X = 4) = {}^{4}C_{4} p^{4} q^{4 - 4} = {}^{4}C_{4} {(\frac{1}{5})}^{4} {(\frac{4}{5})}^{4 - 4} = {}^{4}C_{4} {(\frac{1}{5})}^{4} {(\frac{4}{5})}^{0} = \frac{1}{625}$

Required binomial probability distribution is given below

$X$ $0$ $1$ $2$ $3$ $4$

$P (X)$ $\frac{256}{625}$ $\frac{256}{625}$ $\frac{96}{625}$ $\frac{16}{625}$ $\frac{1}{625}$

We know that mean of the binomial distribution $= μ = n p$

$∴ Required μ = 4 \times \frac{1}{5} = \frac{4}{5}$

Variance of the binomial distribution $= σ^{2} = n p q$

$∴ σ^{2} = 4 \times \frac{1}{5} \times \frac{4}{5} = \frac{16}{25}$

Which is the required variance.

Five bad oranges are accidentally mixed with $20$ good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.

Important Questions on Probability

Learn and Prepare for any exam you want !

Five bad oranges are accidentally mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.

Important Questions on Probability

Learn and Prepare for any exam you want !

Five bad oranges are accidentally mixed with $20$ good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.