Re: lottery probability
Hello, evanar!
This one is a bit trickier than the usual "Lottery" problems.
A player picks 6 lottery numbers from 1 to 45. Eight numbers are chosen at random:
the first 6 are the winning numbers, the other 2 are supplementary numbers.
Prizes are:
Div. 1: 6 winning numbers
Div. 2: 5 winning numbers & 1 supp.
Div. 3: 5 winning numbers
Div. 4: 4 winning numbers
Div. 5: 3 winning numbers a & 1 supp
Find the number of combinations and the probabilities of each
They used the term "combinations" for good reason:
\(\displaystyle \;\;\) the order of the numbers is not considered.
Let \(\displaystyle W\) = winner, \(\displaystyle S\) = supplementary numbers.
There are: \(\displaystyle \begin{pmatrix}45\\6\end{pmatrix}\) ways to select the 6 W's
\(\displaystyle \;\;\)and there are: \(\displaystyle \begin{pmatrix}39\\2\end{pmatrix}\) ways to select the 2 S's.
Hence, the denominator is: \(\displaystyle D\:=\:\begin{pmatrix}45\\6\end{pmatrix}\cdot\begin{pmatrix}39\\2\end{pmatrix}\) ways to select the eight numbers.
Div. 1: six W's
There is
one way to pick the six W's.
\(\displaystyle P(\text{6W})\:=\:\frac{1}{D}\)
Div. 2: 5 W's & 1 S
There are \(\displaystyle \begin{pmatrix}6\\5\end{pmatrix} \,= \,6\) ways to pick five W's
\(\displaystyle \;\;\) and \(\displaystyle \begin{pmatrix}2\\1\end{pmatrix}\,=\,2\) ways to pick an S.
\(\displaystyle P(\text{5W & 1S})\:=\:\frac{6\,\times\,2}{D}\)
Div. 3: 5 W's (exactly)
There are \(\displaystyle \begin{pmatrix}6\\5\end{pmatrix} = 6\) ways to to pick five W's.
The sixth number must not be the sixth W nor either of the two S's.
\(\displaystyle \;\;\)Hence, there are \(\displaystyle 37\) choices for the sixth number.
\(\displaystyle P(\text{5W}) \:=\:\frac{6\,\times\,37}{D}\)
Div. 4: 4 W's (exactly)
There are \(\displaystyle \begin{pmatrix}6\\4\end{pmatrix}\,=\,15\) ways to pick four W's.
The 5th and 6th numbers must not be either of the other two W's nor either of the two S's.
\(\displaystyle \;\;\)Hence, there are \(\displaystyle \begin{pmatrix}37\\2\end{pmatrix}\,=\,666\) ways to pick the 5th and 6th numbers.
\(\displaystyle P(\text{4W}) \:=\:\frac{15\,\times\,666}{D}\)
Div. 5: 3 W's & 1 S
There are \(\displaystyle \begin{pmatrix}6\\3\end{pmatrix}\,=\,20\) ways to pick the 3 W's.
There are \(\displaystyle \begin{pmatrix}2\\1\end{pmatrix}\,=\,2\) ways to pick the one S.
The 5th and 6th numbers must not be any of the other 3 W's nor the second S.
\(\displaystyle \;\;\)Hence, there are: \(\displaystyle \,\begin{pmatrix}37\\2\end{pmatrix} \,= \,666\) ways to pick the 5th and 6th numbers.
\(\displaystyle P(\text{3W & 1S}) \:= \:\frac{20\,\times\,2\,\times\,666}{D}\)
.
