Question

Glad this site still exists! I have a question on probability that is bothering me. In a promotional effort: customers are encouraged to enter a sweepstakes. Each customer picks 9 numbers between 1-50, inclusive. At the end of the promo period, 9 numbers from 1-50 are drawn, inclusive, and if a customer has 9 numbers matching(without order), the customer wins $5,000,000. What is the probability of the customer winning 5 million dollars? There are other parts to this question but Id rather try myself.

Answer 1

Oh I want to mention that I tried (50!/9!) but that didnt seem right

Answer 2

there are \(\binom{50}{9}\) ways to pick the 9 numbers

Answer 3

if order doesn't matter

Answer 4

Ohh right the choose thing so

50!
----
9!(50-9)!

Answer 5

it would be a miracle if \(\frac{50!}{9!}\) was a probability since that number is certainly not between 0 and 1

Answer 6

the denominator is \[\binom{50}{9}\]
now you have picked a right number up to order
there are \(9!\) ways to rearrange the 9 correct numbers

Answer 7

Ahh ok lol I see so for expected value if I am correct its just this

\[\left(\begin{matrix}50 \\ 9\end{matrix}\right)(5,000,000)\]

Answer 8

Just making sure I am on the right track for this problem

Answer 9

Thanks for the help Satellite73

Answer 10

The expected value is given by:
\[\large E[X]=\frac{5,000,000}{\left(\begin{matrix}50 \\ 9\end{matrix}\right)}\]

Answer 11

Hmmm why division?

Answer 12

Oh wait never mind

Answer 13

Because the probability of winning is given by:
\[\large P(winning)=\frac{1}{\left(\begin{matrix}50 \\ 9\end{matrix}\right)}\]

Answer 14

Because it has a probability of

\[\frac{ 1 }{ 2505433700 }\]

Answer 15

hold the phonen

Answer 16

the probability of winning is not \[\frac{1}{\binom{50}{9}}\]

Answer 17

because you don't have to get the numbers in the exact order

Answer 18

hmm maybe i am wrong

Answer 19

But the property of probability has to between 0 and 1

Answer 20

But the symbol being used is for combinations, not permutations.

Answer 21

Therefore we cannot leave it at just \[\left(\begin{matrix}50 \\ 9\end{matrix}\right)\]

Answer 22

no i was thinking that the numerator is \(9!\) the number of ways you can arrange the 9 numbers but maybe i am mistaken

Answer 23

nvm i am wrong

Answer 24

Its 50 numbers and you choose 9

Answer 25

So putting it another way
\[\large 50C9=\left(\begin{matrix}50 \\ 9\end{matrix}\right)\]
where order does not matter.

Answer 26

yeah you are right sorry

Answer 27

Lol its ok thanks for the help :)

Answer 28

We all have our brain farts

Answer 29

\[\frac{1}{\binom{50}{9}}\] is right

Answer 30

You're welcome :)

Answer 31

Glad to see this site doing well and thanks dropot72 and satalite73 for your help

Answer 32

You're very welcome :)

Answer 33

Keep up the good work