From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs

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»

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