Out of 100 such samples, how would you expect to contain 3 defectives i) using the Binomial distribute 4, ii) Poisson distribution.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : DEC 2015

Using binomial distribution

Probability of defective $(P)=2/10=0.1$

Probability of non defective $(q)=1-0.1=0.9$

$n=20$ and $N=100$

$P(x=X)= \space ^nC_x\times p^x\times q^{n-x}$

To expect 3 defective

$P(x=3)= \space ^{20}C_3 (0.1)^3 (0.1)^{17}=0.19$

$\therefore $ No. of packets containing 3 defective

$$= N \times P (X=3)$$

$$ = 100\times 0.19= 19 ... packets$$

2) By poisons distribution

we have $m=2., n=20 , p=0.1\\ P(X=x) =\dfrac {e^{-m}m^x}{x!}$

for 3 defective

$P(X=3)=\dfrac {e^{-2}2^x}{3!}\\ =\dfrac {0.135\times 8 }6 \\ =0.18$

$\therefore$ No. of packets containing 3 defective is $=N \times P(X=3)\\ =100\times 0.18 \\=18 packets.$