I have a twist to my previous probabilty question.?

Question

I have a friend trying to figure out a probability problem with animal measurements.

Say you have 18 mini horses that are measured in inches, (1/4" increments). How would you calculate the probability that 9 of the 18 were 28 inches tall? The height range is 27" - 28" (measured in 1/4" increments - 5 slots).

Thanks in advance for your help!

Answer 1

Notice the following
1. There are a fixed number of horses (18).
2. Each horse's height is independent of the others.
3. There are two categories that we are considering here and for each category, their probabilities do not change. In other words:
Success = a horse is 28" tall and Probability of success = 1/5
Failure = a horse is NOT 28" tall and Probability of failure = 4/5

From these criteria, we can conclude that the problem results in a BINOMIAL distribution.

This means that if we want to calculate the probability a certain number of successes (i.e. exactly 9 horses are 28" tall), we would use the binomial probability formula, which states:

P(x) = (n!/[(n-x)!x!])*(p^x)(q^(n-x))
where:
n = number of horses (in this case, 18)
x = number of successes (in this case, 9)
p = probability of success (in this case, 1/5)
q = probability of failure (in this case, 4/5)

Substituting everything into our formula, we get that:

Probability of exactly 9 horses that were 28" tall
= P(9)
= (18!/[(18-9)!9!])*(1/5)^9*(4/5)^(18-9)
= (18!/9!9!)*(1/5)^9*(4/5)^9
= (18!/9!9!)*(4^9/5^18)
= 48620*(0.0000000687)
= 0.00334

So you have a 0.334% chance of this happening.

Answer 2

( 18! / (9! 9!) ) (4^9 / 5^18 ) should give you the odds of exactly 9 mini horses measuring at 28 inches. Or roughly 1 in 300. The odds of "at least" 9 mini horses measuring at 28 is roughly 1 in 235.

To compute the latter case, add all the terms from n = 9 to n = 0:

Sum (18! / ((18-n)! n!) ) (4^n / 5^18 )