SOLUTION: In a 6/48 lottery game a player selects six numbers from 1 to 48. What is the probability of selecting at least five of the six winning numbers? (Round answer to seven decimal plac

Question 1007053: In a 6/48 lottery game a player selects six numbers from 1 to 48. What is the probability of selecting at least five of the six winning numbers? (Round answer to seven decimal places.)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
i believe this will be equal to probability of getting exactly 5 of the numbers correct plus the probability of getting exactly 6 of the numbers correct.

the general formula is:

p(x) = p^x * q^(n-x) * nCx

nCx = n! / ((n-x)! * x!)

p = 1/48
q = 1 - 1/48 = 47/48

in your problem:

p(5) = p^5 * q^1 * 6C5
p(6) = p^6 * q^0 * 6C6

trsnalate to numbers and you get:

p(5) = (1/48)^5 * (47/48)^1 * 6 = 2.305694078 * 10^(-8)
p(6) = (1/48)^6 * (47/48)^0 * 1 = 8.17622013 * 10^(-11)

add them up and you get p(5 or 6) = 2.313870298 * 10^-8

that is equivalent to .00000002313870298

round that to 7 decimal places and you get .0000000 = 0

here's the printout of the excel worksheet where i did all the calculations.

the total probbility equals 1 as it should.

the probability of getting 5 right or 6 right is the sum of p(x) for x = 5 and 6.

the probability is 0 when rounded to 7 decimal places.

$$$

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