I really ned help with this math question.

1. What are your chances of winning a lottery of 45 numbers if 7 numbers wins it?

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- Dec 18th 2006, 06:45 PMLuvProbability winning lottery
I really ned help with this math question.

1. What are your chances of winning a lottery of 45 numbers if 7 numbers wins it? - Dec 18th 2006, 07:10 PMThePerfectHacker
Probability is the ratio of favorable to possible.

The possible outcomes are all the possible numbers that you can have is the number of different numbers from 45. That is,

$\displaystyle {{45}\choose 7}=1724425560$.

The favorable outcomes are all the possible ways of choosing a win, which is just one. Thus it is 1 in that. - Dec 18th 2006, 07:23 PMLuvStill need help
Thank you very much.

I was just wondering if you could tell me the steps how to do that problem because I'm still confused how you got that answer.

I will very much appreciated it.

Thank you in advanced. - Dec 18th 2006, 07:29 PMThePerfectHacker
That symbol I used (the paranthesis) the number of combinations. Over here it represents the number of ways selecting 7 objects from 45. You use the following formula,

$\displaystyle \boxed{ \frac{45!}{7!(45-7)!} }$

Where the factorial, means, the product of all numbers before it.

For example,

$\displaystyle 5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$

Factorials, get large, quickly.

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Thus, if you wanted to know what would be for 8 out of 45 then it would be,

$\displaystyle {{45}\choose 8}=\frac{45!}{8!(45-8)!}$

And now just use a super calculator (like the one on your computer). - Dec 19th 2006, 04:58 AMCaptainBlack
The probability that the first number chosen is one of yours is 7/45

(as you have 7 of the 45 numbers)

If the first number was one of yours the prob that the second is one of

yours is 6/45 (as you have 6 of the remaining 45 numbers)

If the first two numbers were yours the prob that the third is one of

yours is 5/44 (as you have 5 of the remaining 44 numbers)

and so on untill:

If the first six numbers were yours the prob that the seventh is one of

yours is 1/39 (as you have 5 of the remaining 39 numbers).

The final probability is the product of these:

7! (45-7)! / 45!=1/45379620

RonL - Dec 19th 2006, 05:00 AMCaptainBlack
- Dec 19th 2006, 06:01 AMThePerfectHacker
- Dec 19th 2006, 08:53 AMCaptainBlack