- #1

zmalone

- 10

- 0

Ex:

1. 5

2. 0

3. 0

4. 0

5. 9

6. 8

7. 1

8. 0

9. 8

10. 2

Using a binomial distribution to find the probability of any 3 out of 10 I got:

1/10 chance of getting zero = p(0) = 0.1

P(X = 3) = (10 C 3) * 0.1^3 * (1-0.1)^(10-3) = .057395

This does not take into consideration the idea of 3 in a row though so my next idea was just to simply take the 1/10 odds and cube them to find the odds of getting 3 in a row as 1/1000.

From there, I added the 1/1000 probablilities 8 times assuming random 1-3, 2-4, 3,-5, 4-6, 5-7, 6-8, 7-9, and 8-10 all had a 1/1000 probability.

Adding these 8 "clusters" would result in the probability of 8/1000.

Another idea was using these 8 clusters in place of the 10 choose 3 in the binomial dist.:

P(X = 3) = 8 * 0.1^3 * (1-0.1)^(10-3) = .00382637

PS this is for the prob. of EXACTLY 3 zeros out of 10 numbers being in a row, my next task is to find prob of AT LEAST 3

Kind of confused on which route to take, any help is appreciated. Thank you!