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प्रश्न
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
उत्तर
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(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
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