Question

3 defective bulbs are mixed with 7 good ones .let X be the no. of defective bulbs when 3 bulbs are drawn at random .Find the mean and variance.

Answer 1

heya dear,

here izz your answer in the pic....

also iam really happy to help uh sir

Answer 2

defective bulbs = 3 , good bulbs = 7 , total number of bulbs =7+3=10

since 3 bulbs are drawn at random from 10 bulbs.

let X denote the random variable number of defective bulbs (out of the three bulbs drawn).

then X can take the values 0,1,2 or 3.

P(X=0)=P(getting no defective bulb)=



P(X=1)=P(getting 1 defective bulb and 2 good bulbs)=

P(X=2)=P(getting 2 defective bulb and 1 good bulb)=

P(X=3)=P(getting all the 3 defective bulbs i.e. no good bulb)=

∴ the probability distribution of the discrete random variable X is



X 0 1 2 3
P(X)


now the mean of the distribution



therefore mean =9/10=0.9

X P(X)
0 -0.9 0.81 0.2363
1 0.1 0.01 0.0053
2 1.1 1.21 0.2118
3 2.1 4.41 0.0368
=0.4902


variance=0.4902