Suppose you have a litter of mice which consists of 5 males and 3 females. If you randomly grab two of them without replacement. 1. what is the probability that they are both male?
2. what is the probability that you get at least one female
3. what is the probability that one is male and one is female

Problem 1

  5/14

Explanation:

We have 5 males and 3 females, so there are 5+3 = 8 mice total. The probability of selecting a male is 5/8. After that male is chosen, there are 5-1 = 4 of them left out of 8-1 = 7 total. This subtraction happens because the first mouse selected is not replaced. The probability of picking another male is 4/7.

Multiplying these fractions leads to the answer

(5/8)*(4/7) = (5*4)/(8*7) = 20/56 = 5/14

Note: to reduce 20/56 into 5/14, you divide both parts by the GCF 4.

Problem 2

 9/14

Explanation:

The events "selecting two males" and "selecting at least one female" are complementary events. One or the other, but not both, must happen.

If we let

A = event of selecting two malesB = event of selecting at least one female (ie one or more females)

then we can say

P(A) + P(B) = 1

P(B) = 1 - P(A)

P(B) = 1 - (5/14)

P(B) = (14/14) - (5/14)

P(B) = (14-5)/14

P(B) = 9/14

Note how P(A) = 5/14 is the result from problem 1.

Problem 3

 15/28

Explanation:

The probability of selecting a male is 5/8. After the male is chosen, there are 8-1 = 7 mice left. The probability the second mouse is female is 3/7 since there are 3 female mice out of 7 left total.

We get this result after multiplying the values:  (5/8)*(3/7) = 15/56

Now consider that the first mouse is female. The probability of this is 3/8. The probability that the second mouse is male is 5/7.

The fractions multiply to (3/8)*(5/7) = 15/56. We get the same result as before due to the numerators swapping places, but not much else happens. So we have a sort of symmetry going on here.

The last step is to add the two results we got:

(15/56)+(15/56) = (15+15)/56 = 30/56 = 15/28

Or you could compute it like this

2*(15/56) = (2*15)/56  = 30/56 = 15/28

The '2' is to signify there are two ways to select exactly one male and exactly one female, where the order doesn't matter.

Here's another way we can get the answer.

The probability we get both males is 5/14 found back in problem 1.

The probability we get both females will follow the same idea as problem 1, and we would compute (3/8)*(2/7) = 6/56 = 3/28

Add those fractions:  (5/14) + (3/28) = (10/28) + (3/28) = 13/28

Now consider the events

C = event of selecting two mice of the same gender (both are male OR both are female; pick 1 scenario only)D = event of selecting two mice of different genders (the first is male, the other is female, or vice versa)

Events C and D are complementary events.

We calculated that P(C) = 13/28, which means

P(D) = 1 - P(C)

P(D) = 1 - (13/28)

P(D) = (28/28) - (13/28)

P(D) = (28-13)/28

P(D) = 15/28

There are seven groups of stars and 49 stars in total. 49/7=7 7 stars per group is the answer