# Basic probability question (1 boy + 2 girls of a group of 7)

• MathMan2022
In summary: In this case, the two events are the selection of a team of 1 boy and 2 girls, and the selection of a team of 3 girls and 2 boys. So you are adding the probabilities of these two events: P(1 boy and 2 girls selected) = (C(2,1)*C(5,2))/C(7,3) = 20/35P(3 girls and 2 boys selected) = (C(5,2)*C(7,3))/C(15,6) = 16/35
MathMan2022
Homework Statement
Lets image following problem.

We have a group of 7 students, where you are suppose to select a team of 1 boy and two girls. And that 5 of the team 3 girls and 2 are boys. What is the probability that a team consist of one boy and 2 girls?
Relevant Equations
k*l = number of combinations

K(n,r) = n!/r!*(n-r)!
I know this problem can be done as follows.
P(1 boy and two girls) = (C(2,1)*C(5,2))/C(7,3) = 20/35

My question can this be written as probility fractions? Meaning

Lets say if that (2/5*3/7+1/2*1/7)/(3/7) But that doesn't give same result? what am I doing wrong?

MathMan2022 said:
We have a group of 7 students, where you are suppose to select a team of 1 boy and two girls. And that 5 of the team 3 girls and 2 are boys. What is the probability that a team consist of one boy and 2 girls?
You have a group of 7.
You want to form a team of 3.

But what does this mean: "And that 5 of the team 3 girls and 2 are boys"
Do you mean "5 of the group are 3 girls and 2 boys"?
???

.Scott said:
You have a group of 7.
You want to form a team of 3.

But what does this mean: "And that 5 of the team 3 girls and 2 are boys"
Do you mean "5 of the group are 3 girls and 2 boys"?
???
Sorry for confusion.

I meant to write the following, the 7 students consists of 5 girls and 2 boys. What is the probability of selecting a group of one 1 boy and 2 girls.

Using combination I can deduce that: C(2,1)*C(5,2))/C(7,3) = 20/35

But my question is I attemped also to calculate the result using fractions (see my original post). But can't get the same result? What am I doing wrong?

Writing meaningless fractions, as far as I can see. Where do you get those numbers from?

Explain your reasoning. We can't tell you what you're doing wrong if we don't know what you're doing.

vela
MathMan2022 said:
Homework Statement: Lets image following problem.

We have a group of 7 students, where you are suppose to select a team of 1 boy and two girls. And that 5 of the team 3 girls and 2 are boys. What is the probability that a team consist of one boy and 2 girls?
Relevant Equations: k*l = number of combinations

K(n,r) = n!/r!*(n-r)!

I know this problem can be done as follows.
P(1 boy and two girls) = (C(2,1)*C(5,2))/C(7,3) = 20/35

My question can this be written as probility fractions? Meaning

Lets say if that (2/5*3/7+1/2*1/7)/(3/7) But that doesn't give same result? what am I doing wrong?
An alternative method using direct probabilities is first to calculate the probability that you get BGG, which is:$$p(BGG) = \frac 2 7 \times \frac 5 6 \times \frac 4 5 = \frac 4 {21}$$You can then note that there are three likely equally ways to get one boy and two girls: BGG, GBG, GGB. So, the total probability of getting one boy and two girls is$$p = 3\times p(BGG) = \frac 4 7$$Which agrees with the answer you got by counting.

Last edited:
FactChecker
The first solution is exactly what the probability written as a fraction is.
The second solution is a bad attempt to get the result using probabilities written as a fraction.

MathMan2022 said:
I know this problem can be done as follows.
P(1 boy and two girls) = (C(2,1)*C(5,2))/C(7,3) = 20/35
I hope that you understand why this works. This is all multiplication and there is no addition in this formula
MathMan2022 said:
My question can this be written as probility fractions? Meaning

Lets say if that (2/5*3/7+1/2*1/7)/(3/7) But that doesn't give same result? what am I doing wrong?
You are adding two probabilities together. I think that is wrong. What two events do those probabilities come from? When you add the probabilities of two events, you need to be sure that you are not double-counting their intersection.

Last edited:
berkeman

## What is the probability of selecting 1 boy and 2 girls from a group of 7?

To find the probability, we need to know the composition of the group. Assuming the group has an equal number of boys and girls (3 boys and 4 girls), the probability can be calculated using combinations. The number of ways to choose 1 boy from 3 is C(3,1), and the number of ways to choose 2 girls from 4 is C(4,2). The total number of ways to choose 3 children from 7 is C(7,3). The probability is (C(3,1) * C(4,2)) / C(7,3) = (3 * 6) / 35 = 18/35 ≈ 0.514.

## How do you calculate combinations for probability problems?

Combinations are calculated using the formula C(n, k) = n! / (k!(n-k)!), where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. For example, C(4, 2) = 4! / (2!(4-2)!) = 6.

## What assumptions are made in this probability problem?

The main assumptions are that the group consists of a specific number of boys and girls (e.g., 3 boys and 4 girls) and that each child is equally likely to be chosen. Additionally, we assume that the selections are made without replacement.

## Can the probability change if the group composition is different?

Yes, the probability will change if the number of boys and girls in the group is different. For instance, if there are 2 boys and 5 girls, the calculations for the combinations and the resulting probability would be different.

## What is the total number of ways to choose 3 children from a group of 7?

The total number of ways to choose 3 children from a group of 7 is given by the combination C(7, 3). Using the formula for combinations, C(7, 3) = 7! / (3!(7-3)!) = 35.

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