What is the probability of winning the jackpot for the fifth time on the twentieth attempt?

Question

Slot machines have three drums with the same number of positions on each drum. Each position contains a symbol. When you pull the lever on the machine, the drums spin. Assume that each drum is equally likely to stop in any position, and that the positions in which the three drums stop are independent of each other and independent from spin to spin. Suppose a particular slot machine has 9 positions on each drum, and that each position contains one of two symbols: an eggplant or a grapefruit. You win the jackpot if all three drums stop with eggplants showing. The first drum has 6 eggplants and 3 grapefruits, the second has 5 eggplants and 4 grapefruits, and the third drum has 1 eggplant and 8 grapefruits.

Answer 1

So here are the major parts of this problem: calculating a probability of success p (not a common task in a statistics class since it's typically provided for you) and applying the negative binomial distribution (since you're finding a certain number of successes upon reaching the last of a certain number of trials).

Let's calculate p first. Each drum is independent of each other, and we're looking for three drums coming up on eggplant for success. So this probability is p = (6/9)(5/9)(1/9) = 30/(9³) = 30/729 = 10/243

So p = 10/243. This is half the battle. Now, we apply the negative binomial distribution.

The negative binomial pmf is defined/parameterized in a number of ways, but I'll use this representation:

f(x) = P(X = x) = _(x-1)C_(r-1)*(1-p)^x-rp^r

This applies for two possible outcomes (in our case, jackpot or not), independent trails (in our case, one spin of the drums won't affect the next spin of the drums), and p remaining the same across trials (in our case, each spin of the drums has the same probability p of getting a jackpot).

We are looking for the probability of r = 5 successes (jackpots), with the fifth (and last) jackpot being on the 20th attempt, so x = 20. To be clear on X, it is the number of trials until the r^th success.

So we have p = 10/243, 1-p = 1 - (10/243) = 233/243, x = 20, r = 5.

Plug it in!

P(X = x) = _(x-1)C_(r-1)*(1-p)^x-rp^r

P(X = 20) = _(20-1)C_(5-1)*(1-(10/243))^20-5(10/243)⁵

P(X = 20) = ₁₉C₄*(233/243)¹⁵(10/243)⁵ <--- exact

P(X = 20) ≈ 0.00024, or about 0.024%. <----- approximate

I hope this helps!