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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
Solution
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) = nCxpxqn-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
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