# Math Help - Binomial problem

1. ## Binomial problem

A company establishes a fund of 120 from which it wants to pay an amount, C, to any of its 20 employees who achieve a high performance level during the coming year. Each employee has a 2% chance of achieving a high performance level during the coming year, independent of any other employee. Determine the maximum value of C for which the probability is less than 1% that the fund will be inadequate to cover all payments for high performance.

I'm a little confused as to what to do. The correct answer is $C=60$

Basically I did this:

$P[X=x]={{20}\choose{x}}*.02^{x}*0.98^{(20-x)}<0.01$

Trying to find the integer value of x that gives the solution closest to 0.01 without exceeding it.

I came up with $P[X=3]=0.006469$

So, $C=120/3=40$

This, however, is incorrect and I'm not sure why. Any help?

2. ## Re: Binomial problem

Hey downthesun01.

Just a thought: should you consider all probabilities less than or equal to some number (i.e. P(X <= 0), P(X <= 1), P(X <= 2) etc)?

3. ## Re: Binomial problem

Originally Posted by chiro
Hey downthesun01.

Just a thought: should you consider all probabilities less than or equal to some number (i.e. P(X <= 0), P(X <= 1), P(X <= 2) etc)?

I don't think so.

$P[X=0]=0.6676\text{, }P[X=1]=0.2725\text{, and }P[X=2]=0.0528$

So, those probabilities alone are much larger than 1%.

4. ## Re: Binomial problem

Just out of curiosity with regard to the fund, what is the probability for the fund paying out for each individual?

I'm curious because the probability for fund pay-out for one person might not be the same as the probability of getting a promotion.

5. ## Re: Binomial problem

Each individual has a 2% chance of getting the bonus independent of any other individual getting the bonus. That's why we can use a binomial distribution.

This is what I really need to do:

$P[CN>120]<0.01\text{, where N is the number of individuals that get the bonus, and C is the bonus amount paid to an individual}$

So, I'm trying to find the largest value for C, such that the probability of the initial 120 bonus money allotment not being enough is less than 1%.

6. ## Re: Binomial problem

Suppose that you set the bonus at 120, then you would be in trouble if 2 or more employees qualified.
The probability of that happening is 1 - (0.6676 + 0.2725) = 0.0599.

If you set the bonus at 60, (allowing for 2 qualifiers), your in trouble if 3 or more qualify.
The probability of that is 1 - (0.6676 + 0.2725 + 0.0528) = 0.0071.

That's less than 1% so you are okay setting it at 60.

7. ## Re: Binomial problem

Thank you. I don't why I didn't see it that way.