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 exactly three 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(exactly three defectives) = P(X= 3)

P(x = 3) = `10""_3 (1/20)^3 (19/20)`

= `(10 xx 9 xx 8)/(1 xx 2 xx 3) xx 1/(20)^3 xx (19)^7/(20)^7`

= `120 xx (19)^7/(20)^10`

= `(120 xx 8.945 xx 10^8)/(1.023 xx 10^13)`

= `(120 xx 8.945)/(102300)`

= `1073.4/102300`

= 0.0105

Let x = (19)²

log x = 7 log(19)⁷

= 7 log 19

= 7 × 1.2788

= 8.9516

x = Anti log 8.956

= 8.945 × 10⁸

Put y = (20)¹⁰

log y = (10) log 20

= 10 × 1.3010

= 13.010

y = Anti log 13.010

= 1.023 × 10³