It has 15 things affghhjj ibhcff
Ten thousand is the place value of the 9 in that number
Answer:
P(Y=1|X=3)=0.125
Step-by-step explanation:
Given :
p(1,1)=0
p(2,1)=0.1
p(3,1)=0.05
p(1,2)=0.05
p(2,2)=0.3
p(3,2)=0.1
p(1,3)=0.05
p(2,3)=0.1
p(3,3)=0.25
Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)
P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)
P(X=3)=p(3,1) +p(3,2) +p(3,3)
P(X=3)=0.05+0.1+0.25=0.4
Hence P(Y=1|X=3)=0.125
Answer:
0.070
Step-by-step explanation:
Y = number on trial
Y has a negative binomial distribution
r = 3
P = 30% = 0.3 probability of positive indication.
P(Y = 11) probability of 11 employees that must be tested to get 3 positives
Y-1Cr-1*p^r*q^(y-r)
Y-1 = 11-1 = 10
r-1 = 3 -1 = 2
10C2 x 0.3³x0.7⁸
45x0.027x0.05764801
= 0.070
This is the probability that 11 employees must be tested to get 3 positives.
Sad to say it is likely D. If you are in the United States, I wouldn't know what deductions are available, but here are some possibilities.
1. Gladys is a single Mom. She gets to deduct her child.
2. Gladys owns her own home and gets to deduct her municipal tax. Michelle is renting and may be able to deduct something but not as much.
3. Gladys gets to deduct medical expenses. Michelle does not.
4. Gladys has a travelling allowance that is deductible. Michelle does not.
5. Gladys goes to church and tithes. Michelle does not.
6. Gladys has a registered savings plan. Michelle does not.
The problem is that the two women might very well be in a different tax bracket when all the deductions are considered. That depends on how the US system works. I don't think you are supposed to choose A. All other things being equal, they should be in the same tax bracket.
I don't see how B would come about. Usually state is dependent on Federal (it is in Canada anyway).
C is definitely wrong unless the savings plan is registered. Any savings plan that produces dividends or interest that is not registered is taxable.
