Statistics math problem

Question

The Arizona pick six lotto lottery says you are a winner if your lottery ticket has 3,4,5 or 6 of the winning numbers of 42 to choose from. If you buy one ticket what is the probability of your ticket is a loser?

Answer 1

Pat,

 

I attended Morris Hills Regional HS in Rockaway, NJ, many years ago and lived in Wharton.  Small coincidence!

 

Since the lottery process is assumed randomized and independent, the probability of matching any one of the 42 number is 1/42. We should use the Binomial Distribution to compute the winning probabilities of 3, 4, 5 and 6 matches, and finally subtract their sum from 1 to get the probability of having a losing ticket.

 

Let p = 1/42 = 0.02381 be the probability of success.  Then (1 - p) = 0.97619 is probability of failure.

     n  = number of successful matches on a single ticket

     N  = 42 = total # of numbers that can be matched

 

Binomial: p(n,N) = probability of n successful numbers out of the N numbers

    

                         = {N!/[(N-n)! n!]} pⁿ (1 - p)^(N-n).  where n! denotes the factorial product n(n-1)(n-2)...1

So:

P(3,42) = {42!/[(39)! 3!]} (0.02381)³ (0.97619)³⁹ = {(42)(41)(40)/(3)(2)}(1.34983 x 10-⁵)(0.390697) = 0.06054

P(4,42) = {42!/[(38)! 4!]} (0.02381)⁴ (0.97619)³⁸ = {(42)(41)(40)(39)/(4)(3)(2)}(3.21394 x 10^-7)(0.40022639) = 0.01440

P(5,42) = {42!/[(37)! 5!]} (0.02381)⁵ (0.97619)³⁷ = {(42)(41)(40)(39)(38)/(5)(4)(3)(2)}(0.76524 x 10-⁸)(0.40999) = 0.00267

P(6,42) = {42!/[(36)! 6!]} (0.02381)⁶ (0.97619)³⁶ = {(42)(41)(40)(39)(38)(37)/(6)(5)(4)(3)(2)}(1.82204 x 10^-10)(0.41999) = 0.00040

 

Summing these 4 probabilities = 0.07801 which is having some success with your one ticket.

So NO success probability = 1 - 0.07801  =  0.922

 

Note, on average 92 out of every 100 lottery tickets you bought would be losers.