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.

Answer 1

1/15 is the ans i think

Answer 2

Answer:

Mean of Distribution = 1.3

Step-by-step explanation:

Probability of defective bulb = $\frac{5}{15}\:=\:\frac{1}{3}$

Probability of good bulbs = $1-\frac{1}{3}\:=\:\frac{2}{3}$

Let, X be the no of defective bulb in 4 drawn bulbs.

P [ X=0 ] = P [ 0 defective bulb ] = $\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$

              = $\frac{16}{81}$

P [ X=1 ] = P [ 1 defective bulb ] = $4\times\frac{1}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}$

              = $\frac{32}{81}$

P [ X=2 ] = P [ 2 defective bulb ] = $6\times\frac{1}{3}\times\frac{1}{3}\times\frac{2}{3}\times\frac{2}{3}$

              = $\frac{24}{81}$

P [ X=3 ] = P [ 3 defective bulb ] = $4\times\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}\times\frac{2}{3}$

              = $\frac{8}{81}$

P [ X=4 ] = P [ 4 defective bulb ] = $\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}$

              = $\frac{1}{81}$

Probability Distribution Table is attached.

Means, E(X) = $\sum_{i=0}^{4}x_1p(x_i) = 0\times\frac{16}{81}+1\times\frac{32}{81}+2\times\frac{24}{81}+3\times\frac{8}{81}+4\times\frac{1}{81}$

                     $=0+\frac{32}{81}+\frac{48}{81}+\frac{24}{81}+\frac{4}{81}$

                     $=\frac{108}{81}$

Therefore, Mean of Distribution = 1.3