Answer: 1/5, 1/2, 0.
Step-by-step explanation:
given data:
no of cameras = 6
no of cameras defective = 3
no of cameras selected = 2
Let p(t):=P(X=t)
p(2)=m/n,
m=binomial(3,2)=3!/2!= 3
n=binomial(6,2)=6!/2!/4! = 15
p(3)= 3/15
= 1/5.
p(1)=m/n,
m=binomial(6,1)*binomial(2,2)=6!/1!/4!*2!/2!/0!= 7.5
n=binomial(6,2)= 15
p(2)= 7.5/15
= 1/2
p(0)=m/n,
m=0
p(0)=0
Answer:
a.
The probability of X
k P(X=k)
0 0.25
1 0.5
2 0.25
b. The expected variable value of X; E(X) = 1
Step-by-step explanation:
Given that:
number of cameras = 6
numbers of defective = 3
the probability of defective camera p = 3/6 = 0.5
sample size n = 2
Then X = {0,1,2}
Suppose X is the given variable that represents the number of defective cameras in the sample.
∴
X [tex]\sim[/tex] Bin (n =2, p = 0.5)
The probability mass function of binomial distribution can be computed as :
[tex]P(X =x) = (^n_x) p^x (1-p)^{n-x}[/tex]
For ;
x = 0
The probability P(X=0) [tex]= (^2_0) 0.5^0 (1-0.5)^{2-0}[/tex]
[tex]P(X=0) = \dfrac{2!}{0!(2-0)!} \times 1 \times 0.5^2[/tex]
[tex]P(X=0) =1\times 1 \times 0.25[/tex]
[tex]P(X=0) = 0.25[/tex]
For :
x = 1
The probability P(X=1) [tex]= (^2_1) 0.5^1 (1-0.5)^{2-1}[/tex]
[tex]P(X=1) = \dfrac{2!}{1!(2-1)!} \times 0.5^1 \times 0.5^1[/tex]
[tex]P(X=1) =2 \times 0.5 \times 0.5[/tex]
[tex]P(X=1) =0.5[/tex]
For :
x = 2
The probability P(X=2) [tex]= (^2_2) 0.5^2 (1-0.5)^{2-2}[/tex]
[tex]P(X=2) = \dfrac{2!}{2!(2-2)!} \times 0.5^2 \times 0.5^0[/tex]
[tex]P(X=2) =1 \times 0.5^2 \times 1[/tex]
[tex]P(X=2) =0.25[/tex]
The probability of X
k P(X=k)
0 0.25
1 0.5
2 0.25
The expected variable value of X can be computed as:
E(X) = np
E(X) = 2 × 0.5
E(X) = 1
