Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a universe of $25$ numbers. At the time of the drawing you take $15$ numbers (non-repeatable)

I pick $15$ numbers.

The chances of having $15$ hits is the same to have exactly $5$ ?

Example:

Universe: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25
My picks are: 1,2,3,4,5,6,7,8,9,10,13,14,15

Drawing: 1,5,6,7,8,10,12,13,14,16,18,20,22,24,25

What are the odds of hit $15$ numbers What are the odds of hit $5$ numbers (but only $5$, not more)

share|improve this question
    
You pick 15 numbers and the chances of having 15 hits....what's a hit here? –  DonAntonio Aug 16 '13 at 14:40
    
sorry my native is not english. A hit is having the number picked in the drawing. In the case of the example, I picked number 1 and the draw have number 1 so I have a hit there. –  Bart Aug 16 '13 at 14:41

1 Answer 1

up vote 3 down vote accepted

There are $\binom{25}{15}$ possible draws, only one of which gives you $15$ hits, so the probability of getting $15$ hits is the very small number

$$\frac1{\binom{25}{15}}=\frac1{3,268,760}\approx0.000000306\;.$$

A draw that gives you exactly $5$ hits must contain $5$ of your $15$ numbers and all $10$ of the numbers that you did not choose; there are $\binom{15}5$ such sets, one for each set of $5$ of your choices. The probability that you will get exactly $5$ hits is therefore

$$\frac{\binom{15}5}{\binom{25}{15}}=\frac{3003}{3,268,760}\approx0.000918697\;.$$

This is still small, but it’s much bigger than your chance of getting $15$ hits.

share|improve this answer
    
Thank you very much! –  Bart Aug 16 '13 at 14:43
    
@Bart: You’re very welcome. –  Brian M. Scott Aug 16 '13 at 14:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.