A few days ago
Lucid_dreams

# In lotto 54 a bettor selects six numbers for 1 to 54 (without repetitiion)……?

and a winning six number combination is later randomly selected.

What is the probability of:

1.all six winning numbers

2.exactly 5 of the winning numbers

3.exactly three of the winning numbers

4.no winning numbers..

A little confused about this one it is supposed to use Hypergeometric distribution and i dont understand

A few days ago
darty

1.54*53*52*51*50*49=18595558800

combinations

so 1/18595558800

that is the first one

Edit- sorry i made a mistake it should

actually be 25827165 possibilities

because i forgot to divide it by 720 because

that is 6 factorial.

the second one is

89677.65625 or 25827165 / 288

the third one is

74.662248496762257169287696577243

or 25827165 / 345920

and the last one is

2.1046440731997817383872500796968

or 25827165 / 12271512

the formula is

C(n, k)

P = ————————–

C(k, m) * C(n-k, k-m)

P is possibilities

C is the combinations button on your calculator

it might look like nCr

the n(54) is how many numbers altogether

the k(6) numbers picked out

the m(0-6) is the number of winning numbers

all of the answers are how many possibilities

just do 1/ number of bossibilities

to find the probability

1 day ago – Edit – Delete

Source(s):

http://www.saliu.com/oddslotto.html

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