An elementary school class contains 5 boys and 4 girls. Two students are selected. What is the probability that one girl and one boy is chosen?

Question

An elementary school class contains 5 boys and 4 girls. Two students are selected at random to erase the blackboard. What is the probability that one girl and one boy is chosen?

Answer 1

Michael F.'s answer is correct. He used "counting" formulas to get the answer. It is always nice to be able to check your answer by a different technique, though.

Another approach to this problem is to do it directly, using conditional probabilities. The probability of doing A first, and then B afterward, is P(A)×P(B|A), where the 2nd is the conditional probability of doing B after A has been done. [NOTE: a lot of incorrect answers are obtained when forgetting to use the conditional probability for the second factor!]

So the probability of selecting G, a girl first, is P(G) = 4/9. That would then leave 3 girls and 5 boys, so the probability of selecting a boy second after a girl, P(B|G), is 5/8. Then the probability of doing both (in that order), is found by multiplying 4/9 ×5/8 = 5/18.

In a similar manner, the probability of selecting a boy first and then a girl second after the boy would be

5/9×4/8 = 5/18.

Since the event of selecting one boy and one girl could happen either way, we would have to add the two probabilities to get 5/18 + 5/18 = 5/9 for the probability of getting one student of each in drawing two.

The answer is 5/9, as seen in the answer from the other approach with the Combination formulas.

The counting formulas for combinations and permutations come in handiest when dealing with larger sets. This problem is done just as easily by either approach (as long as done correctly).

Answer 2

Jason has a bad that contains 4 blue marbles, 13 green marbles and 3 red marbles. Jason selects a marble and places it back in the bag. If jason selects a marble 80 times, how many times would you expect him to select a blue marble

 

Answer 3

There are ₉C₂ = 36 ways to select a pair of students from a group of 9.

There are ₅C₂ = 10 ways to select 2 boys, plus ₄C₂ = 6 ways to select 2 girls.

It seems to me that the probability of two of the same sex is (10 + 6)/36 = =16/36 = 4/9

So that the probability of two of opposite sex is 1 - 4/9 = 5/9.

 

There are ₉C₂ = 36 ways to select a pair of students from a group of 9.

There are precisely 5 × 4 = 20 pairs where one member is a girl, 1 member is a boy.

So the probability is 20/36 = 5/9

Answer 4

Of course, as we are not told otherwise, we assume that there is an equal chance for any of the 9 children, whether girls or boys, in any combination or order, to be chosen.

In this scenario, we are choosing 2 students out of 9 & want to know the probability of one being a girl & the other a boy. As 4 of the 9 are girls, there must be a 4/9 probability of a girl being chosen. There being 5 boys, the probability of choosing a boy is 5/9. We now use these fractions to find the probability of choosing a girl & a boy:

4/9 x 5/9 = 20/81

You should always check to see if the resulting fraction can be simplified, but this one cannot. So, there is a slightly less than 1/4 probability of a girl & a boy being chosen.