SOLUTION: I am trying to calculate the odds similar to lottery with different conditions and don't know how to do it. I am not trying to win the lottery. It is for a board game variation s
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Question 576945: I am trying to calculate the odds similar to lottery with different conditions and don't know how to do it. I am not trying to win the lottery. It is for a board game variation similar to Bingo.
The conditions are: 45 numbers, pick 5, 15 numbers randomly chosen. How do I calculate the odds of getting all 5 numbers being drawn? Would I continue with setting up the as 1/45, 1/44, 1/43, 1/42, 1/41 and then 1/40? or would I go 3/45, 3/44....? Would I still use 120 as the dividing factor, or would I divide by 360? Help would be appreciate. Thank you
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
You want 5 successes out of 5 trials where the probability of success on any given trial is 15/45 = 1/3, so the odds are:
\ =\ \left(5\cr 5\right\)\left(\frac{1}{3}\right)^5\left(\frac{2}{3}\right)^0)
But since
\ =\ 1\ \forall\,n\,\geq\,0\,\in\,\mathbb{Z})
and
this reduces to:
\ =\ \left(\frac{1}{3}\right)^5\ =\ \frac{1}{243})
Which is the probability, but you asked for the odds. Odds in favor are the probability of success divided by the probability of failure (which is 1 minus the probability of success), so the odds are:
}{1\ -\ P_5\left(5,\frac{1}{3}\right)}\ =\ \frac{\ \frac{1}{243}\ }{\ \frac{242}{243}\ }\ =\ \frac{1}{242})
1 in 242, or 242 to 1 against.
John
My calculator said it, I believe it, that settles it
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