Answer to Question #125000 in Statistics and Probability for gxngs

Question

Suppose that a box contains 7 cameras and that 5 of them are defective. A sample of 3 cameras is selected at random (without replacement). Define the random variable
X
as the number of defective cameras in the sample.

Write the probability distribution for
X
.

Answer 1

Solution:

Let p(t):=P(X=t)

p(3)=m/n,

m=binomial(5,3)=5!/3!/2!=10,

n=binomial(7,3)=7!/3!/4!=35,

p(3)=10/35=2/7,

p(2)=m/n,

m=binomial(5,2)*binomial(2,1)=5!/2!/3!*2!/1!/1!=10*2=20,

n=binomial(7,3)=35,

p(2)=20/35=4/7,

p(1)=m/n,

m=binomial(5,1)*binomial(2,2)=5!/1!/4!*2!/2!/0!=5,

n=binomial(7,3)=35,

p(2)=5/35=1/7,

p(0)=m/n,

m=0

p(0)=0

Answer:

p(0)=0,

p(1)=1/7,

p(2)=4/7,

p(3)=2/7