What is the probability of winning the lottery if the balls number 1 to 50 and you have to get 5 correct?

Question

What is the probability of winning the lottery if the balls number 1 to 50 and you have to get 5 correct?
Just need help setting up these types of problems and answering them to get me started, thanks for any help or explanations

Answer 1

Probability equals the number of correct answers divided by the number of possible answers .... so for the first ball there are 5 correct numbers from a possible 50 or 5/50=0.1 For the next ball you have only 4 correct ones left out of a possible 49 or 4/49=0.0816. Next is 3/48= 0.0625. Next is 2/47=0.0426. Next is 1/46=0.0217.
Probability of getting all of them you multiply all the answers together so =0.1*0.0816*0.0625*0.0426*0.0217=4.723 times 10 to power negative 7.
Therefore the probability of only the five numbers coming up and no others is .0000004723 Now if you take the reciprocal (which means one divided by that number) of that number you get 2,117,298 and we more commonly say your chances of winning are 1 in 2,117,298 or a little over 2 million to one.

Answer 2

I am assuming that once a number is drawn, it can not be drawn again. For example if a 23 is drawn then the second number must be drawn from the remaining 49 balls.

Then for the first number the possibilities are 1in 50 (1:50) for the second (1:49), third (1:48), fourth (1:47) and fifth (1:46).
Then the odds are 50 * 49 * 48 * 47 * 46 = 254,251,200
i.e. one chance in 254,251,200. However this assume syou must draw the numbers in a particular order. If the numbers can be random, (drawn in any order) then the number of possible combinations for the 5 balls is 5! (factorial) which is
5 *4 *3 * 2 *1=120, so divide 254,251,200 by 120 and get
(1:2,118,760)

reference: http://icarus.mcmaster.ca/fred/Lotto/

Moral: don't play the lotto. It is a tax for people who don't understand statistics.

Answer 3

You did not mention getting 5 out of 6 correct numbers. Therefore the draw is 5 numbers and the order does not matther.

the anwer is the chances of getting 5 numbers right x number of way to get it right (e.g the winning number is 1,2,3,4, 5. How many ways you can arrange it. If my ticket is 2,3,5,1,4 I still win, so is 1,5,3,2,4)

Your first anwerer gave you the chances of getting 5 numbers right. You should figure out how many ways to arrange 5 non-repeating numbers; someone has answered that part too.

An alternative way to set it up would be
prob(hitting 1 of the 5 winning numbers from 50) x prob(hitting 1 of the remaining 4 winning numbers from 49) x ..... prob(hitting 1 of the last 1 number from 46).
5/50 x 4/49 x 3/48 x 2/47 x 1/46. This way you don't have look at the permutation part. (also this set up works well if the question is draw 6, and you need to hit 5 numbers..)

Hope you enjoy your math class. I did.

Answer 4

I believe it's

(1/50) x (1/49) x (1/48) x (1/47) X (1/46)

or a bit over 1 in 250 million.

Answer 5

Assuming that the order the numbers are chosen doesn't matter, then it is a combination problem and you use the formula (n choose k) = n! / (k!(n-k)!) where n is the number of items and k is the number chosen. In this case n=50 and k=5.

It's a big number. Be good at math, put your money in a nice mutual fund and let all the people who are bad at math throw their money away on the lottery.

Answer 6

i think its to do with factorials

50! = how many possible outcomes

50! x 49! x 48! x 47! x 46!