Probability question/s

Question

A class at College is comprised of sixteen (16) girls and fourteen (14) boys. The class was given detention from 3:00 pm to 4:30 pm last week Tuesday. At 3:30 pm the teacher selected a student at random and allowed the student to go home.

At 4:00 pm the teacher selected another student at random and also allowed that student to go home.

Determine the probability that gender of the students chosen were:

(i) Both male 3

(ii) The same gender

(iii) Different genders

(iv) At least one male

Answer 1

Starting off, we should consider all different outcomes of this scenario. The breakdown of all possible scenarios is:

1. Student 1 M, Student 2 M
2. Student 1 M, Student 2 F
3. Student 1 F, Student 2 M
4. Student 1 F, Student 2 F

Using this information, we can break down the various parts of the question and add the respective probabilities together as such:

1. Both Male (Case #1)
2. The same gender (Case #1 + Case #4)
3. Different genders (Case #2 + Case #3)
4. At least one male (Case #1 + Case #2 + Case #3)

With this in mind, we can look towards solving the probability for each individual case:

*Note: Remember that once a child is selected to go home, they are out of detention and cannot be selected again.

1. Student 1 M, Student 2 M
2. 14/30 * 13/29 = 182/870
3. Student 1 M, Student 2 F
4. 14/30 * 16/29 = 224/870
5. Student 1 F, Student 2 M
6. 16/30 * 14/29 = 224/870
7. Student 1 F, Student 2 F
8. 16/30 * 15/29 = 240/870

So, when we look back on the questions and add the respective probabilities we get:

1. Both Male (Case #1)
2. 182/870, or 91/435
3. The same gender (Case #1 + Case #4)
4. 182/870 + 240/870 = 422/870, or 211/435
5. Different genders (Case #2 + Case #3)
6. 224/870 + 224/870 = 448/870, or 224/435
7. At least one male (Case #1 + Case #2 + Case #3)
8. 182/870 + 224/870 + 224/870 = 630/870, or 21/29

As a bit of an added bonus, you can note that the sum of Case #1-4 adds up to be 1 (100%, since it covers all possible cases). With this in mind, you could calculate the probabilities the question asks for as such:

1. Both Male (Case #1)
2. The same gender (Case #1 + Case #4)
3. Different genders (Case #2 + Case #3) -> (1 - (Case #1 + Case #4)) -> (1 - [Question #2])
4. At least one male (Case #1 + Case #2 + Case #3) -> (1 - Case #4)

This is called solving the probability using the inverse, and is quite common throughout more complex probability questions.