00:01
In this question it stated that 35 % of the population has blue eyes.
00:05
So that means for any randomly selected person from the population, this probability is 0 .35, and for parts a and b, we consider a random selection of 6 people, and we're interested in the probability that exactly 3, and that 3 or 4 have blue eyes out of the sample of 6.
00:27
So let's first define a random variable x as the number of people in the sample of six who have blue eyes.
00:34
Here, each of the six people can be thought of as bernoulli trials with two outcomes of interest, either have blue eyes or not, and since the people are randomly selected, their outcomes are independent.
00:44
So the number of successes in a fixed number of independent bernoulli trials is a binomial random variable.
00:52
So here, x is a binomial based on six trials and probability success.
00:58
0 .35.
01:00
The probability function for the binomial random variable is given by this formula.
01:17
For part a, we want the probability that 3 have blue eyes.
01:21
So this is the probability that x is equal to 3.
01:25
Using the probability function that 6 choose 3 times 0 .35 to the exponent 3, times 0 .65 to the exponent 3.
01:39
And this comes out to approximately 0 .2355.
01:49
And then for b, we want the probability that 3 or 4 have blue eyes, probability that 3 have blue eyes, plus the probability that 4 have blue eyes.
02:15
So we solve the probability that 3 have blue eyes in part a, that's .2355.
02:24
And for 4, we have 6 choose 4 times .35 to the exponent 4 times .65 to the exponent 2, and this comes out to .3306.
02:53
And then for part c we consider a group of size 100.
03:00
These 100 are randomly selected, and we're asked if 40 people with blue eyes is unusually high.
03:09
So we have in this group 40 with blue eyes...