1=51
2=34
3= 95 (i think)
4=38
The answer to your question,
If rounded it would be 12.3, if not 12.5 with a repeated.
-Mabel
This problem can be solved from first principles, case by case. However, it can be solved systematically using the hypergeometric distribution, based on the characteristics of the problem:
- known number of defective and non-defective items.
- no replacement
- known number of items selected.
Let
a=number of defective items selected
A=total number of defective items
b=number of non-defective items selected
B=total number of non-defective items
Then
P(a,b)=C(A,a)C(B,b)/C(A+B,a+b)
where
C(n,r)=combination of r items selected from n,
A+B=total number of items
a+b=number of items selected
Given:
A=2
B=3
a+b=3
PMF:
P(0,3)=C(2,0)C(3,3)/C(5,3)=1*1/10=1/10
P(1,2)=C(2,1)C(3,2)/C(5,3)=2*3/10=6/10
P(2,0)=C(2,2)C(3,1)/C(5,3)=1*3/10=3/10
Check: (1+6+3)/10=1 ok
note: there are only two defectives, so the possible values of x are {0,1,2}
Therefore the
PMF:
{(0, 0.1),(1, 0.6),(2, 0.3)}
Answer: -1>4
Step-by-step explanation:
