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

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Homework Help Overview

The problem involves calculating the probability of winning a lottery by selecting 5 different numbers from a total of 45 available numbers. The order of selection does not matter, and the inquiry is focused on determining the probability of matching all 5 drawn numbers with the chosen numbers using only one ticket.

Discussion Character

  • Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the formula for combinations to calculate the probability of winning the lottery. They express uncertainty about the correctness of their method and the resulting probability.

Discussion Status

Some participants seek clarification on the original poster's question regarding the correctness of their method and answer. There is a brief exchange about the calculations presented, with one participant affirming the correctness of the approach without further elaboration.

Contextual Notes

The original poster edited their post to clarify their question and correct errors, indicating an ongoing process of refinement in their understanding of the problem.

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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?
 
Last edited:
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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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