Question

If 5% of the items produced turn out to be defective, then find out the probability that out of 20 items selected at random there are atleast two defectives

Answer 1

Probability of getting a defective item

p = `5/100 = 1/20`

q = 1 – p

⇒ q = `1 - 1/20`

= `(20 - 1)/20`

q = `19/20` and n = 10

In binomial distribution

P(X = x) = nC_xp^xq^n-x 

Here (X = x)= `10"C"_x (1/20)^x (19/20)(10 - x)`

p(atleast two defectives)

= p(x ≥ 2)

= p(x = 2) = p(x = 3) + ………….. + p(x = 10)

= 1 – p(x < 2)

= 1 – {p(x = 0) + p(x = 1)}

= `1 - [10"c"_0 (1/20)^0 (19/20)^(10 -0) + 10"C"_1 (1/20)(19/20)^(10 - 1)]`

= `1 - {(19)^10/(20)^10 + (10 xx 1/20) + (19)^9/(20)^9}`

= `1 - {(19)^9/(20)^10 + (10(19)^9)/(20)^10}`

= `1 - {(19)^9/(20)^9 [19 + 10]}`

= `1 - [(3.229 xx 10^11 (29))/(1.023 xx 10^13)]`

= `1 - [(3.229 xx (29))/(1.023 xx 10^2)]`

= `1 - [93.641/102.3]`

= `1- 0.9154`

= 0.0846

Let x = (19)⁹

log x = 9 log 19

= 9 × 1.2788

= 11.5092

x = Anti log 11.5092

= 3.229 × 1011