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

[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?

 

[.Scott]

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"?
???

 

[MathMan2022]

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?

 

[mjc123]

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.

 

[PeroK]

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.

 

[Gavran]

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.

 

[FactChecker]

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

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.