Question

From a lot of 15 bulbs which include 5 defectives, 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

Let the number of defective bulbs be represented by a random variable X. X may have value 0, 1, 2, 3, 4. 

If p is the probability of getting defective bulb in a single draw then p = 5/15 = 1/3

∴ q = Probability of getting non defective bulb = 1 - p = 1 - 1/3 = 2/3

Since each trial in this problem is Bernoulli trials, therefore we can apply binomial distribution as

P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...n

P(X = r) = ⁴C_r(1/3)^r(2/3)^{4 - r}

Now, P(X = 1) = ⁴C₁(1/3)¹(2/3)³ = 4 x 1/3 x 8/27 = 32/81

P(X = 2) = ⁴C₂(1/3)²(2/3)² = 6 x 1/9 x 4/9 = 24/81

P(X = 3) = ⁴C₃(1/3)³(2/3)¹ = 4 x 1/27 x 2/3 = 8/81

P(X = 4) = ⁴C₄(1/3)⁴ = 1/81

Now probability distribution table is

X	0	1	2	3	4
P(X)	16/81	32/81	24/81	8/81	1/81

Now mean E(X) = ∑p_ix_i = 0 x 16/81 + 1 x 32/81 + 2 x 24/81 + 3 x 8/81 + 4 x 1/81

= (32 + 48 + 24 + 4)/81 = 106/81 = 4/3