Question

There are six girls and eight boys on a high school debate team Two students from the team are chosen at random to attend a debate workshop. What is the probability that at least one of the students chosen is a girl?

Answer 1

How do I solve these types of problems?

Answer 2

Let g be the event a girl is chosen and b the event that a boy is chosen

To have at least one girl in 2 tries you can have

gg
bg
or
gb

Where gg means a girl was chosen first and second and bg means a boy was chosen first and a girl chosen second

Do yo see that?

Answer 3

Okay, but how do I find the probability that a girl is selected at least once?

Answer 4

On the first try what is the probability that a person chosen is a girl?

You have 6 girls and 8 boys, right? What fraction of these are girls?

Answer 5

There are 14 total people, so there is 6 chances a girl is chosen out of 14. So 6/14?

Answer 6

correct!

So for the second selection, you will not have 14 people, but 13 because the girl was chosen on the first slot. So now there will be 5 girls and 8 boys right?

At this point, what is the chance of choosing a girl?

Answer 7

5/13 but why would you assume a girl was chosen rather than a boy on the first pick?

Answer 8

Because we are saying, what if we choose a girl on the first then on the second. We will take care of the case where a boy was selected first next.

Now, what we have done is figured out the chance of gg, which is
$$
\cfrac{6}{14}\times\cfrac{5}{13}
$$

Let's take the next case, bg. What is the chance that the 1st choice is a boy? We have 6 girls and 8 boys at this point.

Answer 9

So 6/14 again

Answer 10

No, 8/14 - do you see why? This is the chance of a boy being selected first

What we are doing is computing mutually exclusive cases. All cases are

gg,gb,bg,bb, right? So the probability, P(gg) + P(gb) + P(bg) + P(bb) = 1.

Do you see that? We just care about the case where there is at least one girl. that's only 3 of the 4 cases. In fact you can almost see the answer without all these computations. I'll show you that shortcut in a sec.

Let's compute P(bg):

$$
\cfrac{8}{14}\times\cfrac{6}{13}
$$

Do you see why?

Answer 11

Okay, I think I understand now.

Answer 12

Next, what is P(gb)? That's the last one

Answer 13

It would be 6/14 * 8/13

Answer 14

Perfect!

Now let's all three results and see what we get...

Answer 15

Now let's add all three results and see what we get ...

Answer 16

$$
\cfrac{6}{14}\times\cfrac{5}{13}+\cfrac{8}{14}\times\cfrac{6}{13}+\cfrac{6}{14}\times\cfrac{8}{13}=?
$$

Answer 17

9/13 (So 69%?)

Answer 18

Yes!

Answer 19

Here is the easier way...

Answer 20

We know that P(gg) + P(gb) + P(bg) + P(bb) = 1

So P(gg) + P(gb) + P(bg) = 1- P(bb)

This means we really just needed to compute P(bb).

Do you see that?

Answer 21

$$
P(gg) + P(gb) + P(bg) = 1- P(bb)=1-\cfrac{8}{14}\times\cfrac{7}{13}
$$

Answer 22

And this equals 9/13.

Much easier approach

Answer 23

Okay thank you! That makes a lot more sense, and will make it easier to solve future problems. :)

Answer 24

You're welcome