In a school, 70% of the students are female. 10 students were randomly selected.

Question

In a school, 70% of the students are female. 10 students were randomly selected. What is the probability that?

1. Five female students are selected

2. At least are one female student is selected

3. Three males are selected

Answer 1

The first thing you want to establish is that your outcome is dichotomous, male or female. Where P(Female) = .7 and P(Male) = .3, with n = 10. If we assume that the probability of a female or male being selected is constant (i.e. if the first person selected of the 10 is female the probability the next person is female would not change) then we can treat this information like it comes from a binomial distribution, where random variable X = number of female selected out of 10 trials where P(female) = .7. X~B(10,.7)

P(X=x) = (n Choose x) * P(Female)^x * P(Male)^(n-x)

where (n Choose x) = n!/(x!*(n-x)!)

where ! represents the factorial function i.e. n! = n*(n-1)*(n*2)*(n-3)*......(n-(n-2))*(n-(n-1))

So now we can answer part A;

where part A is asking what is the probability of 5 females being select of the 10 total people who get selected.

So P(X = 5)

= (10 Choose 5)*(.7^5)*(.3^(10-5))

= (10!/(5!*(10-5)!))*(.7^5)*(.3^5)

= 252*0.168*0.00243

= 0.1029 = probability exactly 5 women are selected for the random sample of 10 people.

This works because the choose function gives use the number of possible combinations that you can make up a 10 person group where there are 5 females and 5 males, P(Female)^5 is the probability of their being 5 females on any single group of 10, and P(Male)^5 is the probability of their being 5 males on any single group of 10 people.

So you can think of it this way the probability the first 5 people are female and the last 5 are male would be .7*.7*.7*.7*.7*.3*.3*.3*.3*.3 = (.7^5)*(.3^5) and that is one possible way it could happen, so we multiply (.7^5)*(.3^5)*(n Choose x) because (n Choose x) is all the possible ways we could make a group of 10 people with 5 being female and 5 being male. We can use this same type of thinking for the next two questions as well.

For part B)

At least 1 female is selected or P(X>=1) this would require use to do P(X=1)+P(X=2)+P(X=3)+)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10),

but because we know that our sample space ranges from 0 to 10 (the fewest number of females in the group can be 0 and the most can be 10) and that if we added up the probability of all the possible outcomes of X they would equal the value of 1

i.e. P(X=0)+P(X=1)+P(X=2)+P(X=3)+)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10) = 1

so the short cut would be P(X>=1) = 1-P(X=0)

P(X=0) = (10 Choose 0)*(.7^0)*(.3^10)

= 1*1*0.0000059049

= (5.9049E^-6)

So P(X>=1) = 1-(5.9049E^-6) = .999994095 ≈ 1

Part C.

3 males are select of the 10 so that is the same as saying 7 females are selected, Then we can plug the values into our equation

P(X=7) = (10 Choose 7)*(.7^7)*(.3^3)

= 120*0.08235*0.027

= 0.26683