# Probability Calculation - Lottery

1. Feb 22, 2010

### Hart

1. The problem statement, all variables and given/known data

To win a lottery, must pick 5 different numbers from the 45 available.

The order in which the numbers are chosen does not matter.

With only one ticket, what is the probability of winning (i.e. matching all 5 numbers drawn with all 5 chosen) ?

2. Relevant equations

Stated within the solution

3. The attempt at a solution

n = number of elements in the field (in this case, 45)
p = number of choices (5)

$$P(win) = \left(\frac{n!}{(p!(n - p)!)}\right)$$

Therefore:

$$=\left(\frac{45!}{(5!(45- 5)!)}\right)$$

$$=\left(\frac{45!}{(5!)(40!)}\right)$$

$$=\left(\frac{45!}{(120)(40!)}\right)$$

$$=\left(\frac{45!}{(5!(40)!)}\right)$$

$$= \left(\frac{45*44*43*42*41}{120}\right)$$

$$=1221759$$

Therefore:

$$P(win)= \left(\frac{1}{1221759}\right) \approx 8.18\times10^{-7}$$

.. Is this correct method / answer?

Last edited: Feb 22, 2010
2. Feb 22, 2010

### kuruman

$$\frac{45!}{(5!)(40!)}=\frac{45*44*43*42*41*40!}{5!*40!}=\frac{45*44*43*42*41}{5!}$$

Is there a question here?

3. Feb 22, 2010

### Hart

.. Just edited my initial post to make it a bit clearer and to correct some errors. Wanted to know if this was the correct method for calculating the probability?, and hence the correct answer?

4. Feb 22, 2010

### kuruman

It is correct.
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