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 [tex]p=\dfrac{6}{100}=0.06[/tex]
Binomial probability distribution:-
[tex]P(X=x)=^nC_xp^x(1-p)^{n-x}[/tex]
We have ,
a)
[tex]P(X=0)=^{10}C_0(0.06)^0(0.94)^{10}\\\\=(1)(0.94)^{10}\\\\=0.538615114095\approx0.5386[/tex]
b)
[tex]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[/tex]
Thus , [tex]P(X>2)=0.0188[/tex]
