Megamillions Probability - 3 of 5 numbers Quoted directly from the state gambling website:

Players must pick six numbers from two
  separate pools of numbers - five different numbers from 1 to 56, and one number from 1 to 46.
  You win the jackpot by matching all six winning numbers in the "random drawing"

I was trying to figure out what the probability would be for you to draw 3 of the 5 numbers from the first pool, and draw a number from the second pool. In other words, correctly get $\frac{3}{5}$ numbers and the megaball.
The website claims its $\frac{1}{13,781}$ but I was wondering how they got to that figure and whether or not it's accurate
 A: $\dfrac{5*4*3*51*50}{56*55*54*53*52} * {5 \choose 3} * \dfrac{1}{46} = \dfrac{1}{13781}$
Probability of selecting 3 correct and 2 incorrect from the first pool, with the appropriate binomial coefficient, AND (multiply) also getting the one from the second pool correct.
A: The number $X$ of numbers that you get correct is a hypergeometric random variable with parameters $N=56$ (population size), $K=5$ (the matching numbers, that is the numbers that you have selected) and $n=5$ (sample size, i.e. numbers drawn). So, the probability of matching $3$ numbers is equal to $$P(X=3)=\frac{\dbinom{5}{3}\dbinom{51}{2}}{\dbinom{56}{5}}=0,003337857111$$ Now, you also need to match the megaball and the probability for this is similarly equal to $$\frac{\dbinom{1}{1}\dbinom{45}{0}}{\dbinom{46}{1}}=\frac{1}{46}$$ Since, these two drawings are independent you multiply their probabilities to find the requested probability, which is equal to $$0,003337857111\cdot\frac{1}{46}=0,000072562111$$ while $$\frac{1}{13781}=0,000072563675$$ so the required probability is almost equal (rounded) to $\frac{1}{13781}$
