I got this question as my country is conducting a lottery this weekend.
So my country's lottery works like this:
A person buys a ticket with $6$ numbers from a pool of $45$ numbers, $1 – 45$. During the draw, the gamemaster will draw $6$ numbers. Afterwhich, he will draw 1 more number from the remaining $(45 – 6 = ) \space 39$ numbers and calls this the additional number.
The top $5$ prizes are as follows:
- 1st Prize – Ticket matches all $6$ numbers
- 2nd Prize – Ticket matches only $5 \space \text{of} \space 6$ numbers AND the additional number
- 3rd Prize – Ticket matches only $5 \space \text{of} \space 6$ numbers
- 4th Prize – Ticket matches $4 \space \text{of} \space 6$ numbers AND the additional number
- 5th Prize – Ticket matches $4 \space \text{of} \space 6$ numbers
The question is: What is the respective probability of the prizes?
I am guessing for the 1st, 3rd & 5th prize respectively
$$
P(\text{1st prize}) = \frac{6 \choose 6}{45 \choose 6} \\
P(\text{3rd prize}) = \frac{{6 \choose 5}{39 \choose 1}}{45 \choose 6} \\
P(\text{5th prize}) = \frac{{6 \choose 4}{39 \choose 2}}{45 \choose 6} \\
$$
Are these calculations correct, and also what is the respective probability of the rest of the prizes?
Best Answer
$$p(1)=\binom 66 / \binom{45} 6 $$
$$P(2)=\binom 65 * \binom {39}1/ \binom{45} 6 * \binom 1 1/ \binom{39} 1 $$ $$P(3)=\binom 65* \binom{39} 1 /\binom{45}6 $$ $$P(4)=\binom 64 * \binom {39} 2 / \binom{45} 6 * \binom 1 1/ \binom{39} 1 $$ $$P(5)=\binom 64 * \binom {39} 2 / \binom{45} 6 $$