Probability of choosing both students are not boy

[desmond iking]

Homework Statement

A class consists of 25 boys and 15 girls . Two students are selected randomly. what is the probability of both students are not boys?

Homework Equations

The Attempt at a Solution

my working is (15C1 X 14C1) /(40C1 x 39C1) = 7/52

or 15/40 x 14/39 = 7/52


but the ans is 8/13

 

[HallsofIvy]

Do you mean "both students are not boys" which is the same as "both are girls" or do you mean "the students are not both boys" which means at "least one is a girl"

In either case, your working is much too complicated. It looks like you are trying to apply formulas rather than thinking about the problem.

Assuming you meant the first then there are originally 40 students, 15 of whom are girls. The probability the first student chosen is a girl is 15/40= 3/8. There are then 39 students, 14 of whom are girls. What is the probability the second chosen is a girl? What is the probability both are girls?

Assuming you mean the second then there are originally 40 students, 15 of whom are girls. The probability the first student chosen is a girl is 15/40= 3/8. There are then 39 students, 25 of whom are boys. What is the probability the second chosen is a boy? What is the probability of "girl, boy", in that order?

Of course, then we have to compute the other order- there are originally 40 students, 25 of whom are boys. The probability the first student chosen is a boy is 25/40= 5/8. There are then 39 students, 15 of whom are girls. What is the probability the second chosen is a girl? What is the probability of "boy, girl", in that order?

What is the probability of "boy, girl" in either order?

(In either case, the answer is NOT "8/13"!)

 

[desmond iking]

Do you mean "both students are not boys" which is the same as "both are girls" or do you mean "the students are not both boys" which means at "least one is a girl"

In either case, your working is much too complicated. It looks like you are trying to apply formulas rather than thinking about the problem.

Assuming you meant the first then there are originally 40 students, 15 of whom are girls. The probability the first student chosen is a girl is 15/40= 3/8. There are then 39 students, 14 of whom are girls. What is the probability the second chosen is a girl? What is the probability both are girls?

Assuming you mean the second then there are originally 40 students, 15 of whom are girls. The probability the first student chosen is a girl is 15/40= 3/8. There are then 39 students, 25 of whom are boys. What is the probability the second chosen is a boy? What is the probability of "girl, boy", in that order?

Of course, then we have to compute the other order- there are originally 40 students, 25 of whom are boys. The probability the first student chosen is a boy is 25/40= 5/8. There are then 39 students, 15 of whom are girls. What is the probability the second chosen is a girl? What is the probability of "boy, girl", in that order?

What is the probability of "boy, girl" in either order?

(In either case, the answer is NOT "8/13"!)

by considering case i and case ii , and then add up the bth probability i got the ans of 8/13 finally.

i misunderstood the question so i ended up getting the probability of both are girls only.

 

[D H]

An easier way to look at it: The complement of the event "both selected students are not boys" is "both selected students are boys". The sum of the probabilities of these two complementary events must be one. The probability that both selected students are boys is easily computed: It's 25/40 * 24/39 = 5/13. The probability that both selected students are not boys is thus 1-5/13, or 8/13.