HW 4, Problem 2, Number 2

[Chelsi Whitely]

I am still confused on how to set up question number two in problem two. I looked at the other threads and the compliment rule, but I am still confused. Would we just do 1-(1/9) since the probability of a child having high levels is just 1/9 chances?

[Taeho Kim]

Hi Chelsi,

Try to follow the steps and let me the part you are stuck.

i) Write down all the possible outcomes of this binomial random variable.
It should be {0, 1, 2, 3, ..., 14, 15}.
Again, these are possible values X can have.

ii) Then what is the event we are interested in?
It is {X>1}, right?
Now, circle all the outcomes which satisfy the condition X>1.
It should be part of all the possible outcomes above.

iii) If we define it A, i.e., A={X>1}.
Then, what is the A complement?
Can you think of the outcomes in the A complement?
Those are simply the outcomes outside of the circle in ii).

iv) If you can find out the outcomes,
Pr[A complement] should be simple from the binomial formula.


v) Lastly, Pr[X>1] = Pr[A] = 1 - Pr[A complement].


 

[Chelsi Whitely]

Ok so I have 14 possible outcomes for A and 2 possible outcomes for A compliment. But I'm confused at the part using the binomial formula. I know n=15 and so does j=14 and p=1/15? I don't understand where to get p for the formula and if these numbers are right.

[Taeho Kim]

Okay. We are almost there.

{A complement} = {X<=1} = {X=0} or {X=1}.

What you have to do is using the formula twice.
Take a look at the formula carefully and find out the two j's
you have to plug in to the formulas.

[Chelsi Whitely]

So, j=0 one time doing the formula and j=1 the other time doing the formula? And p=1-A complement which is: p=1-0.9333=0.066 since A=14/15=0.933. Is all this correct?

[Chelsi Whitely]

Oh wait. p=1-A compliment. A compliment is 2/15=0.1333. So, then p=1-0.1333=0.866?

[Taeho Kim]

So, j=0 one time doing the formula and j=1 the other time doing the formula?
=> This is correct.

And p=1-A complement which is: p=1-0.9333=0.066 since A=14/15=0.933. Is all this correct?
=> I am confused this part. Are you claiming that Pr[A complement] is 0.9333?
If this is the addition of the number you got from the formulas, then you are right.

[Chelsi Whitely]

Yes I was saying Pr[A compliment]=14/15=0.933

[Taeho Kim]

What are Pr[Y=0] and Pr[Y=1] through the formula?
Can you show me the result?

[Chelsi Whitely]

Pr[Y=0]=   15!
              --------    (1)(1-0.933)^15-0 = 2.14550926E-7??
              0! (15-0)

[Taeho Kim]

The p was incorrect.

p is the individual success probability.
In this question success is high blood led level.

Then from the problem statement what is the probability that a child has a high blood led level?

[Chelsi Whitely]

I put 1 for p because in the formula I did (0.933)^0 which equaled to 1. So, if that is wrong then I am still confused at how to get p.

[Taeho Kim]

Don't worry, we are about to reach there.

p is "the probability that a child has the high blood lead level" and the problem says:
In a certain population, 1 child in 9 has a high blood lead level

What do you think about p now?
Can you see it?

[Chelsi Whitely]

Pr[Y=0]=   15!
              --------    (1)(1-0.111)^15-0 = 1.49E10??
              0! (15-0)

Is this right?

[Taeho Kim]

The formula is correct.
Check out the number one more time.

[Chelsi Whitely]

I am stumped. I've tried like eight times to figure out what p is and I'm still not getting it.

[Taeho Kim]

Well, you gave me the right p.
Because I can see you plugged in the right number into the formula.
But my question was what is the calculated value from the formula?
It is nothing but (1-0.111)^15, right?
What is the value you obtain from the calculator?

[Chelsi Whitely]

I got 0.171

[Taeho Kim]

Alright. Then similarly what is Pr[Y=1]?
If you add those two, Pr[Y=0] and Pr[Y=1],
then it is Pr[A complement].

Then Pr[A] is immediate.

Please go back to the STEPS again.
Hope you can see the reasoning now.

[Chelsi Whitely]

I think I get it. Thank you for your help