In my last two posts, we looked at ways to make up time with probability questions on the GRE by quickly eliminating answer choices, and then guessing the correct answer. In those posts, we examined the following question:

**From a group of 5 managers (Joon, Kendra, Lee, Marnie and Noomi), 2 people are randomly selected to attend a conference in Las Vegas. What is the probability that Marnie and Noomi are both selected? **

**(A) 0.1**

**(B) 0.2**

**(C) 0.25**

**(D) 0.4 **

**(E) 0.6**

In this post, we’ll examine 2 different ways to find the correct solution to this question.

To begin, it’s important to know that, when it comes to probability questions, we often have 2 distinct approaches to choose from:

- Apply various probability rules
- Apply counting techniques and the standard probability formula

For some questions, it may be best to apply probability rules, and, in other cases, it may be best to use counting techniques. The approach you choose may also depend on your level of comfort with each strategy.

**1. Applying Probability Rules**.

To apply probability rules, I’ll first ask, “**What needs to occur in order for both Marnie and Noomi to be selected?**”

Well, the first person selected must be either Marnie or Noomi AND the second person selected must be the remaining person. At this point, I know that I can apply the AND probability rule to solve this question.

The “AND” probability formula looks something like this: P(A AND B) = P(A) x P(B)

Aside: Please note that there are different ways to represent this formula. One involves using conditional probability, which has some complicated notation. Rather than use this notation, I’ll just add an important stipulation to the above formula. The stipulation is when we calculate P(B), we must assume that event A has already occurred.

Okay, now let’s solve the question.

I know that P(Marnie and Noomi are both selected) = P(one of the two friends is selected first, AND the remaining friend is selected second)

When we apply the formula, we get:

P(M and N both selected) = P(one of the two friends is selected first) x P(the remaining friend is selected second)

Now, what is the probability that one of the two friends is selected first? Well, there are 5 people to choose from, and we want one of the 2 friends to be selected. So, the probability is 2/5 that one of the two friends is selected first.

Now, we need to find the probability that the remaining friend is selected second. Well, first we will assume that one of the friends was already chosen on the first selection. So, at this point there are 4 people remaining, and 1 of those 4 people is the remaining friend. So, the probability is 1/4 that remaining friend is selected second.

This means that P(M and N both selected) = (2/5) x (1/4) = 1/10

So, the correct answer is A.

**2. Applying Counting Techniques**.

To apply counting techniques, we will use the fact that, if we have an experiment where each outcome is equally likely to occur, then

**P(A) = [# of outcomes where A occurs]/[total # of outcomes]**

So, P(M and N both selected)= [# of outcomes where both are selected]/[total # of outcomes]

As I explained in my last post, it’s always best to calculate the denominator first. So, for the denominator in our question, we need to determine the total number of outcomes when 2 people are selected from a group of 5 people. Since the order of the selected people does not matter in the given question, we can apply the combination formula. So, we can select 2 people from 5 people in _{5}C_{2} ways. When we apply the combination formula, we see that _{5}C_{2}= **10**.

At this point, we need to determine the number of outcomes where Marnie and Noomi are both selected. In other words, in how many ways can we select 2 people such that both of those people are Marnie and Noomi?

Well, there’s only **1** way to do this; select both Marnie and Noomi.

So, P(M and N both selected)= **1/10**

Now, some readers may question whether or not there is only 1 way to select both Marnie and Noomi. This is a valid question. After all, we could select Marnie and then Noomi, or we could select Noomi and then Marnie. This would mean that there are 2 ways to select both Marnie and Noomi. Seems reasonable enough. However, the important point to keep in mind is that, when we determined the value of the denominator, we assumed that the order in which the people are selected does not matter. So, when we determine the value of the numerator, we must once again assume the order in which the people are selected. This means that selecting Marnie and then Noomi is the same as selecting Noomi and then Marnie, in which case we cannot consider them as two different outcomes. So, there is only 1 way to select both Marnie and Noomi.

### Most Popular Resources

Can we say that this event is not independent to each other, since we are selecting a person from a different group each time then why are we using the formula of the independent rule?

Hi Jshree,

These are not independent events, since the first person chosen affects the probability of the second person being chosen. In fact, what we are doing is multiplying the probability that one friend is chosen by the probability that the second friend is chosen

given thatthe first friend was already chosen! This is slightly different than the typical “and” rule for independent or dependent events events, which is what Brent mentions in the “Aside” as he is describing the probability method. I encourage you to think about this logically instead of through formulas: in order for both friends to be chosen, we know that one friend must be chosen first (P=2/5). The second choice depends on the first, since there are only four people left. However, since we know how the first choice will affect the second choice, we can find the probability of both being chosen by finding the probability that the second will be chosen given that the first is already chosen (1/4). When we multiply the two together, we get 1/10. If you are still confused, see this website with a good explanation of how to deal with dependent events like this : https://www.mathsisfun.com/data/probability-events-conditional.htmlEven there is another way to get it though . Probability of finding a person from 5 people is 1/5 and to find the second person, the probability changes to 1/4 because we have already selected one person and there are two ways select the person .so the probability to select Marnie and Noomi is

2*(1/5)*(1/4)=0.1

That also does work. The reason we didn’t separately mention that is that— from a strictly numerical standpoint— your method simply adds an extra step to the (2/5)*(1/4) equation. 2*(1/5) = (2/5) and can just be expressed as 2/5.

Still, differentiating between the 1/5 odd that the first person is selected and the 1/4 odds that the second person is selected is a great way to think about that. if adding an extra step like that to similar problems on the GRE helps you think clearly about the way that the odds work, go for it!

Forgive me if this is mentioned elsewhere, but perhaps another quick way of arriving at the answer is to note that selecting one of five people yields a probability of 1/5 = 0.2 and, therefore, since the probability of selecting two predetermined people is less, the answer must be less than 0.2. Of course, this is just taking advantage of the answer choices in this particular question.

I have to admit that, when I first read your solution, I thought that you had found an even faster approach. However, I then realized that the basis of that approach isn’t quite correct.

The problem lies in this statement: selecting one of five people yields a probability of 1/5 = 0.2 and, therefore, since the probability of selecting two predetermined people is less, the answer must be less than 0.2.

To show that this reasoning may be faulty, let’s look at an extreme example. What if 4 people were selected from A, B, C, D, and E and we wanted to know the probability that A, B, C and D are all selected.

If we apply your approach, we would conclude that the probability is less than 0.2, but the probability actually equals 0.2

Even more extreme, what if we selected 5 people from A, B, C, D, and E and we wanted to know the probability that A, B, C, D and E are all selected. Here, the probability is definitely greater then 0.2 (it equals 1)

Cheers,

Brent

I read in a blog about this question related to probability,

Q: There are 8 employees including Bob and Rachel. If 2 employees are to be randomly chosen to form a committee, what is the probability that the committee includes both Bob and Rachel?

I solved it using simple probability approach which is 2/8 * 1/7 = 1/28.

But there was a solution to this question which I could not able to grasp, and also it looks like as if they have solved it considering the position of the person which should not make any difference.

||**************************

We can choose any first person. Then, if we have Rachel or Bob as first choice, we can choose any other person out of 6 people. If we have neither Rachel nor Bob as first choice, we can choose any person out of remaining 7 people. The probability that the committee includes both Bob and Rachel is.

P = 1 – (2\8 * 6\7 + 6\8 * 1) = 2\56 = 1\28

****************************||

Please do provide me some logic in getting this understand.

Hi Joydev,

Your solution is MUCH better.

The logic in the solution shown above basically says:

P(both Bob and Rachel) = 1 – P(not both Bob and Rachel)

= 1 – [P(B or R first AND not B or R second) + P(neither B nor R first AND anyone second)]

= 1 – [(2/8)(6/7) + (6/8)(7/7)]

Cheers,

Brent

Hi Brent

I did this question like this , to select M and N both , we will say : P( M & N ) OR P( N& M)

i.e

Prob ( M and N ) + Prob ( N and M)

(1/5*1/4) + (1/5 * 1/4) = 2/20 =0.1

But I think it will be lengthy for bigger problems/

Hi Tushar! You’re method is valid. Since your calculation assumes that order does matter, you must take into consideration the outcomes M then N and N then M. In this case, the calculation is relatively quick to do. As you noted, though, for a more complicated question, it may take much longer than using combinations as explained in the post 🙂