probability question?

Question

In the MD Lottery game, the palyer chooses six different integers from 1 to 40. If the six numbers match ( in any order) the six different winning number s drawn by the lottery, the player wins the grand prize jackpot, which starts at $1 million and grows weekly until a winner is declared. For each $1 bet, the player must pick two sets of six integers.



For a $1,
a. What is the probability of winning the grand prize jackpot?


b. What is the probability of getting no winning numbers at all?


For two $1 bets
c. WHat is the probability of winning the grand prize jackpot?

Answer 1

The number of ways of choosing 6 numbers from 40 is C(40,6) = 3838380
Only 1 of these gives a winning combination.
So P = 1/3838380 = 2.6e-7

The number of ways you could pick 6 of the 34 numbers which are not in the draw = C(34,6) = 1344904
So P = 1344904 / 3838380 = 0.35

OK for the first bet the prob of not winning =3838379 /3838380
For 2nd bets the prob of not winning on both occasions
= 3838378/3838379
This assumes that the 2 bets are deliberately different.
If they can be the same bet twice, the that value will be 3838379 /3838380 as well.
So Pof not winning on both tries
= 3838379 /3838380 *3838378/3838379
= 3838378/3838380

Answer 2

Probability, eh. (BTB, these answers can probably be found at th MD Lottery website).

a) Let's think about it this way. I have 40 numbers to choose from, but I can only pick six, and there are six winning numbers.

So, for my first number I choose, I have a 6 out of 40 chance of it being a winning number. Right? There are 6 winning numbers and 40 to choose from.

For the second number, I only have a 5 out of 39 chance. Why? Because, now, there are only 5 correct numbers left and only 39 to choose from.

For the third number, I then only have a 4 out of 38 chance of picking a correct number. You see the pattern.

So, the probability of doing multiple things, is simply the probability of each thing multiplied together.

So, the probability of me getting all six numbers would be 6/40 x 5/39 x 4/38 x 3/37 x 2/36 x 1/35 = 1/3838380

or 1 out of 3,838,380

b) No winning numbers?

There are 6 winning numbers. So for me not to get even one, that means there are 34 wrong numbers.

For my first number to be a nonwinning number, I have a 34/40 chance.

For the second, a 33/39 chance...

So, 34/40 x 33/39 x 32/38 x 31/37 x 30/36 x 29/35 = 1 out of 2.854.

That's right. You have better than a 1 out of 3 chance of not getting any numbers whatsoever.

c) Two bets? That's easy, you just double your odds from a). 2/3838380, or 1 out of 1,919,190.

Answer 3

a) 1 / 40C6 = 1 / (40! / 34!6!) = 1 / 383380

b) (34/40) x (33/39) x (32/38) x (31/37) x (30/36) x (29/35) = 0.35ish

c) 2 / 383380, assuming that the two bets are on different sets of numbers in the same week.
Answer would be slightly different if it were one bet in consecutive weeks.
Then it would be 1 - ((383379/383380)*(383379/383380))

Answer 4

a.1/ 40C6
b. 34P6/40P6