finding the probability of winning lottery

Question

i saw an question in stackexchange .it attract me because the given answers are different from one another.Moreover , i think that the accepted solution is wrong and the second most voted answer is correct.I am putting the question in here.

Blockquote Lottery: Finding the probability of winning respective prizes.

What do you think about it. Which solution is correct? I think the the second..

Answer 1

The explanation given in the answer by Satish Thulva (the accepted answer, but strangely now deleted) is correct.

For all $5$ probabilities, the denominator should be ${\large{\binom{45}{6}}}$ since that's the number of possible tickets.

As Satish Thulva explained, there are $3$ groups:

• Group $1$: The first $6$ numbers chosen in the draw.$\\[4pt]$
• Group $2$: The $1$ additional number chosen in the draw.$\\[4pt]$
• Group $3$: The $38$ numbers not chosen in the draw.

hence, of the ${\large{\binom{45}{6}}}$ possible tickets, we get the following counts $$ \begin{array} {|c|c|} \hline \text{prize #}&\;\;\;\text{number of winning tickets}\;\;\;\\ \hline 1&{\vphantom{\Large{\frac{0}{0^2}}}}{\large{\binom{6}{6}}}\\ \hline 2&{\vphantom{\Large{\frac{0}{0^2}}}}{\large{\binom{6}{5}\binom{1}{1}}}\\ \hline 3&{\vphantom{\Large{\frac{0}{0^2}}}}{\large{\binom{6}{5}\binom{38}{1}}}\\ \hline 4&{\vphantom{\Large{\frac{0}{0^2}}}}{\large{\binom{6}{4}\binom{1}{1}\binom{38}{1}}}\\ \hline 5&{\vphantom{\Large{\frac{0}{0^2}}}}{\large{\binom{6}{4}\binom{38}{2}}}\\ \hline \end{array} $$ matching the approach suggested in Satish Thulva's answer.

To explain in a little more detail, consider the count for Prize #$4$ . . .

A winning ticket for Prize #$4$ must have

• $4$ numbers from Group $1$:$\;{\large{\binom{6}{4}}}$ choices.$\\[6pt]$
• $1$ number from Group $2$:$\;{\large{\binom{1}{1}}}$ choices.$\\[6pt]$
• $1$ number from Group $3$:$\;{\large{\binom{38}{1}}}$ choices.

hence by the product rule,$\;\,{\large{\binom{6}{4}\binom{1}{1}\binom{38}{1}}}\;\,$is the number of winning tickets for Prize #$4$.