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:
A's mixture is more acidic.
Step-by-step explanation:
Given:
Two friends mix water and citric acid.
One (Let A) add 3 cups of citric acid for every 16 cups of water and another (Let B)adds 2 cups of citric acid for every 12 cups of water.
Now, we have to find that whose mixture is more acidic means:-
Solution:
A add 3 cups of citric acid for every 16 cups of water, the ratio = 
B adds 2 cups of citric acid for every 12 cups of water = 
Now, to find that who mix more citric acid in the mixture, firstly we will equalize the denominator of both fraction by taking LCM so that can be compare easily.
LCM of 16 and 12:- <u>48</u>
16 - 16, 32, <u>48</u>, 64, 80, 96....
12 - 12, 24, 36, <u>48, </u>60, 72.......
For friend A - 
For friend B - 
<em>We found that in terms of 48 cups of water, friend A mixes 9 cups of acid while friend B mixes 8 cups of acid, means A is adding more acid than B and</em> hence friend A's mixture is more acidic.
5/4=1 2/4
So basically 5/4=6/4
<span>z^2-z(z+3)+3z
z^2 - z^2 - 3z + 3z
0 + 0
0
</span><span>z^2-z(z+3)+3z = 0</span>
6/8 is equivalent to 3/4, and 3/4=.75, and .75/100=75 percent. So yes, it is.