Tough probability question?

Question

This relates to Deal or No Deal, but I have it simplified here:

You have 5 out of 26 cases. There are numbers 1-26 in each of those 26 cases. You want the 9 highest numbers in your cases. You want as many of those 9 numbers as possible. A perfect shootout would be all 5 of your cases containing 5 of those 9. What is the probability that atleast one of your cases contains atleast one of those 9 numbers ?

This is not as easy as it seems. Atleast to me, and I got an A+ in calculus and finite math.

Answer 1

This probability problem can be used using Binomial Distribution.

Given a number of 'x' success, the probability in a binomial experiment is:

P(x) = [n choose x]*p^x*q^(n-x)

where:

n choose x is a combination problem: n!/[(n-x)!x!]
p = probability of success
q = probability of failure
n = total selection
x = successes

In this problem, we are looking for at least one case has a success (which is one of the highest 9 numbers). In essence, we are solving for:

P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(X=5).

However it would be easier to find the probability of the failures and subtract from the total probability because we only have to calculate 1 probability. This would be:

1 - P(x=0) is what we are looking for. Now all we need is to calculate the probabilty.

For P(x=0), we have the following:

n = 5
x = 0
p = (9/26) = probability of success
q = (17/26) = probability of failure

Plugging into P(x), we get:

P(x=0) = (5!/5!)(9/26)^0(17/26)^5

Solving, you get P(x=0) = 0.12

Therefore,

1 - P(x=0) is ------>

1- 0.12 = 0.88

Below is a website devoted to Binomial Distribution exercises
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Hope this helps

Answer 2

The probability that at least one of the five cases contains at least one of the highest numbers is

1 - the probability that all the five cases contain numbers less than 18.

To compute the probability that all five cases contain numbers less than 18, first choose one case containing the numbers 1 through 17. This probability is 17/26. Next choose a second case with the lower numbers. This probability is 16/25. Similarly, choose the remaining three cases with the lower numbers.

The probability of choosing five cases with the lower numbers is the product of the probabilities of choosing each of the five such cases:

17/26*16/25*15/24*14/23*13/22 = .0941

So the probability that at least one case contains at least one of the 9 highest numbers is .9059 or 90.59%.

Answer 3

Prob(at least one of the cases has a number) = 1 - Prob(no cases have those numbers)

So what are the odds of the 5 cases you pick being ones that do not have the numbers you're looking for, or all have one of th 17 numbers you don't want?

=(17/26)(16/25)(15/24)(14/23)(13/22)=0.094071146=probability of getting none of the answers you want

Prob(at least one case has one of the 9 highest numbers=1-0.094071146=0.905928854

Answer 4

Since you have 5 cases and you want the top 9 out of 26 in at least one of the cases then it becomes a hypergeometic probablity

N = 26 total cases
M= 9 number of good cases
n =5 number of cases chosen
k= 1 number of good cases wanted

P(at lease1 good case) = nCk * (N-n)CM/ (NCM)
P(at lease1 good case) = 5C1 * 21C9/ (26C9)
5* (21!/12!*9!)/(26!/(17!9!)
5*(293930)/65780
1469650/3124550
0.47035573


or about 47.03% chance.