Lottery Ticket Question A lottery ticket contains six different numbers, chosen from 1 to 39. The winning ticket will match all six numbers in the correct order, plus a bonus number, which may match the other six numbers. The second prize matches the six winning numbers in the correct order, but not the bonus number. What is the probability of winning first or second prize?
 A: The portion "plus a bonus number, which may match the other six numbers. The second prize matches the six winning numbers in the correct order, but not the bonus number", clearly suggests that the bonus number is separately drawn from the $39$ numbers.
Thus there are two cases:


*

*$\underline{Bonus\; number\; is\; among\; the\; six\; on\; the\; ticket}$


You can win only the first prize, with $Pr = \dfrac{1}{39\cdot38\cdot37\cdot36\cdot35\cdot34}$ 


*

*$\underline{Bonus\; number\; is\; outside\; the\; six\; on\; the\; ticket}$


You can win the second prize, with $Pr = \dfrac{38}{39}\times\dfrac{1}{39\cdot38\cdot37\cdot36\cdot35\cdot34}$
You can win the first prize, with $Pr = \dfrac{1}{39}\times\dfrac{1}{39\cdot38\cdot37\cdot36\cdot35\cdot34}$
A: Edited: just saw that they had to be different numbers
The of winning first or second prize is simply the chance of winning the second prize, or $\frac 1{39*38*37*36*35*34}=\frac 1{2349088560}$ , because the chance of winning first prize is just one of the 39 ways to match six numbers with random bonus numbers.
