Thread: Binomial Distribution

A gambler decides to keep betting on red at roulette, and stop as soon as she has won a total of five bets.

a) what is the probability that she has to make exactly 8 bets before stopping?

b) what is the probability that she has to make at least 9 bets?

Originally Posted by Yan

A gambler decides to keep betting on red at roulette, and stop as soon as she has won a total of five bets.

a) what is the probability that she has to make exactly 8 bets before stopping?

b) what is the probability that she has to make at least 9 bets?

First you need to tell me what's the probability of getting red in a single spin of the wheel.

i don't know, i only got the problem and i don't know how to do it....

Originally Posted by Yan

i don't know, i only got the problem and i don't know how to do it....

Then call the value p. You can chase up the value from your instructor.

Now use the negative binomial distribution: Negative binomial distribution - Wikipedia, the free encyclopedia

the percentage of bets on red each time is (1/2)

Originally Posted by Yan

the percentage of bets on red each time is (1/2)

I'll do it using only the binomial distribution (you might not be familiar with the negative binomial distribution).

a) You need exactly 2 successes in the first 7 trials and then a succes on the 8th trial.

Let X ~ Binomial(n = 7, p = 1/2).

Then Pr(8 trials) = .


b) Pr(9 or more trials) = 1 - Pr(8 or less trials) = 1 - Pr(8 trials) - Pr(7 trials) - ..... - Pr(4 trials) - Pr(3 trials).

Calculate Pr(7 trials), Pr(6 trials), ...... Pr(3 trials) in a similar way to how Pr(8 trials) was calculated.