# Thread: plan of attack

1. ## plan of attack

hi guys,

i have this problem that i have been brewing over for several hours.

Fred's game allows gamblers to select 5 numbers from the numbers 1 to 34. If all 5 chosen numbers are subsequently selected by Fred's random number selector machine, the gambler wins a prize. What is the minimum number of games that the gambler must play so that the probability of winning is greater than 9%?

judging by the marks that it would be allocated (only 2) i assume it must be fairly simple and im simply missing the simplicity... that or it is much harder. i suspected that it may be a binomial probability question but still got no where.

how can i go about solving this problem?

2. Originally Posted by downunder
hi guys,

i have this problem that i have been brewing over for several hours.

Fred's game allows gamblers to select 5 numbers from the numbers 1 to 34. If all 5 chosen numbers are subsequently selected by Fred's random number selector machine, the gambler wins a prize. What is the minimum number of games that the gambler must play so that the probability of winning is greater than 9%?

judging by the marks that it would be allocated (only 2) i assume it must be fairly simple and im simply missing the simplicity... that or it is much harder. i suspected that it may be a binomial probability question but still got no where.

how can i go about solving this problem?
Actualy your question is not well formed. Does Fred's random number generator sample with or without replacement?

RonL

3. so i wasnt the only one wondering about that. im ausuming that they are without replacement though I could be totally wrong

the answer for above is 25044 btw, though i just need to know how to acctaully come to get that number for other variations of the problem

likewise for numbers 1-33 with 6 chosen and a requirment of at least 3% prob the answer is 33228