Question

From a lot of 30 bulbs which include 6 defectives,a sample of 4 bulbs is drawn at random with replacement.Find the probability distribution of the number of defective bulbs.

Answer 1

n(S)=30,A={6 defective bulbs}⇒n(A)=6

⇒ Number of non-defective bulbs = 30 − 6 = 24

4 bulbs are drawn from the lot with replacement. Let X be the random variable that denotes the number of defective bulbs in the selected bulbs.
P(X=0)=P (4 non-defective and 0 defective) ⁴C_o=\(\frac{4}{5}.\frac{4}{5}.\frac{4}{5}.\frac{4}{5}=\frac{256}{625}\)=256/625
P (X = 1) = P (3 non-defective and 1 defective) =  ⁴C₁ \((\frac{1}{5}).(\frac{4}{5})^3=\frac{256}{625}\)
P(X=2)=4C2p2q2P (X = 2) = P (2 non-defective and 2 defective)= ⁴C₂ \((\frac{1}{5})^2.(\frac{4}{5})^2=\frac{96}{625}\)
P (X = 3) = P (1 non-defective and 3 defective)= ⁴C₃  \((\frac{1}{5})^3.(\frac{4}{5})=\frac{16}{625}\)
P (X = 4) = P (0 non-defective and 4 defective)= ⁴C₄  \((\frac{1}{5})^4.(\frac{4}{5})^0=\frac{1}{625}\)
Probability distribution
Therefore, the required probability distribution is as follows.

x	0	1	2	3	4
p(x)	\(\frac{256}{625}\)	\(\frac{256}{625}\)	\(\frac{96}{625}\)	\(\frac{16}{625}\)	\(\frac{1}{625}\)