Help with probability question?

Question

In a lottery, the probability of the jackpot being won in any draw is 1/60. How many consecutive draws need to be made for there to be a greater than 98% chance that at least one jackpot prize will have been won?

The answer is 233 draws but I don't know how to work it out. Any hints?

Answer 1

1/60 = 0.016* (* = repeating)

You can solve the problem by doing it "backwards", as an instance of conditional probability of independent events.  

The probability of no jackpot in trial 1, P1 =s 59/60 = 0.983*

The probability of no jackpot in trials 1 & 2, P2 = (0.983*)2

The probability of no jackpot in trials 1, 2 through n, Pn = (0.983*)^n

All we need to do now is find the value of n such that Pn < 0.02.

And guess what? - P233 = 0.0199

Answer 2

P(at least 1 jackpot) = 1 - P(no jackpot)

P(at least 1 jackpot) > 0.98

=> P(no jackpot) <= 0.02

=> (59/60)^n <= 0.02

=> n ln(59/60) <= ln(0.02)

=> n >= ln(0.02) / ln(59/60) = 232.8 (sign change 'cos ln(59/60) is negative)

=> n = 233

Answer 3

so probability of not getting jackpot in one draw is 1-1/60 = 0.98333333...

so we want to have a more than 98% chance of winning a jackpot in other words less than 2% chance of not getting jackpot

so let x be number of draws and setup inequality

0.9833333..^x < 0.02

natural log both sides

x ln 0.9833333... < ln 0.02

now switch sign because you are dividing by ln 0.9833... which is negative  

x > 232.7598897

so next greatest integer possible is 233 draws

hope this helps