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:
28%
Step-by-step explanation:
Set up a proportion
x / 100 = 7 / 25
Cross multiply
25x = 700
Divide both sides by 25
x = 28
I hope this helped and please mark me as brainliest!
Answer:
Hi! What's your question?
Step-by-step explanation:
The answer is D because I just took my quiz and got it right:)
<span>a becomes 8 times larger when b is doubled.
Let's take the original expression 3b^3/c and effectively double b by multiplying by 2. Then we will take the new expression and divide by the old expression and see what happens.
a = 3b^3/c
a' = 3(2b)^3/c
a'/a = 3(2b)^3/c / 3b^3/c
You can divide by a fraction by swapping the top and bottom and then multiplying. So
a'/a= 3(2b)^3/c * c/3b^3
The c terms cancel, giving:
a'/a= 3(2b)^3/1 * 1/3b^3
The 3's cancel, giving:
a'/a= (2b)^3/1 * 1/b^3
Expand the parenthesis
a'/a= 8b^3/1 * 1/b^3
Cancel the b^3 terms
a'/a= 8/1 * 1/1 = 8 * 1 = 8
So a' is 8 times larger than a.</span>
