What Is the Probability of Winning a 5-Number Lottery with One Ticket?

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Homework Statement



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) ?

Homework Equations



Stated within the solution

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?
 
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<br /> \frac{45!}{(5!)(40!)}=\frac{45*44*43*42*41*40!}{5!*40!}=\frac{45*44*43*42*41}{5!}<br />

Is there a question here?
 
kuruman said:
Is there a question here?

.. 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?
 
It is correct.
 
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