Question

Suppose that a batch of 100 items contains 6 that are defective and 94 that are not defective. If X is the number of defective items in a randomly drawn sample of 10 items from the batch, find (a) P { X = 0 } and (b) P { X > 2 } .

Answer 1

Answer: a)  0.5386     b) 0.0188

Step-by-step explanation:

Let  X is the number of defective items in a randomly drawn sample of 10 items from the batch with parameter :

n= 10 and $p=\dfrac{6}{100}=0.06$

Binomial probability distribution:-

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

We have ,

a)

$P(X=0)=^{10}C_0(0.06)^0(0.94)^{10}\\\\=(1)(0.94)^{10}\\\\=0.538615114095\approx0.5386$

b)

$P(X>2)=1-P(x\leq2)=1-(P(x=0)+P(x=1)+P(x=2))\\\\=1-(^{10}C_0(0.06)^0(0.94)^{10}+^{10}C_1(0.06)^1(0.94)^{9}+^{10}C_2(0.06)^2(0.94)^{8})\\\\=1-(0.5386+(10)(0.06)(0.94)^9+(\dfrac{10!}{8!2!})(0.06)^2(0.94)^8)\\\\\approx1-(0.5386+0.3438+0.0988)\\\\=1-0.9812=0.0188$

Thus , $P(X>2)=0.0188$