Lesson: Introduction to Probability

Comment on Introduction to Probability

is the solution to the last problem done like this?
1/5 (is the chance that he will be selected first) ,
4/5 x 1/4 = 4/20 = 1/5 (is the chance that he will be selected second) ...
1/5 + 1/5 = 2/5 = 4/10 = 0.4
gmat-admin's picture

That's a perfect approach. Since this video is an introduction to probability, we don't get into any techniques yet, other than to explain what probability is all about.

Hi Brent,

This is in reference to the comment made by HellexTV.

Since we need to calculate the number of outcomes where Amir is selected in a 2 person group, can we use combinations here since the order of him being selected first or second does not matter?

So it could be 4C1 ways which is equal to 4. Since Amir has to form a 2 person group with 4 other people?

Please let me know if this approach is correct?

gmat-admin's picture

That approach is perfect. Nice work!

Brent, I've got a hard probability question, and I can't seem to figure out how to tackle it. Would you be so kind to take a look?

"If an integer 'n' is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?"

A: 1/4
B: 3/8
C: 1/2
D: 5/8
E: 3/4

gmat-admin's picture

This question has been discussed many times on the GMAT discussions forums.

You'll find 4 or 5 diverse solutions from experts here: http://www.beatthegmat.com/if-an-integer-n-is-to-be-chosen-at-random-fro...

Thanks brent, much appreciated!

Hi Brent.

For the last question:

The denominator would be 5C2=10
The numerator for the two-person selections including Amir will be say =1*4C1. (fix Amir as one (1) and variate the other person selected (simple combination without order).

So answer would be 1*4c1/5c2 = 4/10.

Now, How this formula will work for below mentioned question?

Two people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir & Brian is selected?

Please tell me where I'm wrong in above mentioned formula?

2 people are fixed, so nominator would be 2x3C0 = 2
and denominator would be 5C2.

Answer would be 2/10. Which is wrong I know because only 1 group can be formed with Amir and Brian.

Where I'm wrong to understand this formula? am I wrong with Denominator?
gmat-admin's picture

Your new question: Two people are randomly selected from the group of five people: Amir, Brian, Claudia, Dhana and Ebo. What is the probability that Amir & Brian are BOTH selected?

The correct answer to that question is 1/10. You're getting 2/10 as your answer, because you're treating the numerator different from the way you're treating the denominator.

For the denominator, you say there are 10 ways to select 2 people from 5 people. You used COMBINATIONS (5C2 to be exact), because you are saying that the ORDER in which we select the two people does not matter. For example, selecting Dhana first and Ebo 2nd, is the SAME as selecting Ebo first and Dhana 2nd. This is a perfectly fine approach.

However, for the numerator, you are used the Fundamental Counting Principle, in which you say there are 2 ways to select BOTH Amir and Brian. This means you are saying that the ORDER DOES matter. That is, you are saying that selecting Amir first and Brian 2nd, is DIFFERENT FROM selecting Brian first and Amir 2nd.

To get the correct answer, we must EITHER decide that order matters OR order does not matter, and then apply that strategy to the numerator AND the denominator.

If we go with your original solution (where the denominator is calculated as though order does NOT matter), then we must also treat the numerator the same way. If order does NOT matter, then we must determine the number of ways to select 2 people from a group of two people (Amir and Brian). Since order does not matter, we'll use combinations. We can select 2 people from 2 people in 2C2 ways. 2C2 = 1

So, P(both Amir and Brian are selected) = 1/10


ALTERNATIVELY, we can treat BOTH the numerator and denominator as though order DOES matter.

So, for the denominator, in how many ways can we select 2 people from 5 people?
We can select the 1st person in 5 ways
Then we can select the 2nd person in 4 ways
So, the TOTAL number of ways to select 2 people (if order DOES matter) = (5)(4) = 20

Then, for the numerator, in how many ways can we select 2 people from 2 people (Amir and Brian) if order DOES matter?
We can select the 1st person in 2 ways
Then we can select the 2nd person in 1 way
So, the TOTAL number of ways to select 2 people (if order DOES matter) = (2)(1) = 2

So, P(both Amir and Brian are selected) = 2/20 = 1/10

Does that help?


Wow! what a quick response. Got the point, great help. Thank you very much.

Hi Brent
Could you solve the following problem below without Combinatorics. I would like to see how you would approach the the problem otherwise, or rather with just simple probability. I can't imagine during a test having to list out all the possible ways one would not get his hat. Is there a faster, more intuitive way?

Al, Bob, Cal and Don each own 1 hat. If the 4 hats are randomly distributed so that each man receives exactly 1 hat, what is the probability that no one receives his own hat?

gmat-admin's picture

Question link: https://gmatclub.com/forum/al-bob-cal-and-don-each-own-1-hat-if-the-4-ha...

If we try to apply probability rules to this question, we quickly become lost in a tangled mess of WHAT-IFS. Here's what I mean:

Let's say P(no one gets their hat) = P(Al doesn't get his hat AND Bob doesn't get his hat AND Cal doesn't get his hat AND Don doesn't get his hat)
= P(Al doesn't get his hat) x P(Bob doesn't get his hat) x P(Cal doesn't get his hat) x P(Don doesn't get his hat)

Since we're examining Al first, we know that P(Al doesn't get his hat) = 3/4
There are 4 hats to start, and 3 of them are NOT Al's hat.
Easy enough.

There are now 3 hats remaining.

What is P(Bob doesn't get his hat)?

If Al received Bob's hat in the first step, then NONE of the remaining 3 hats is Bob's hat, which means P(Bob doesn't get his hat) = 3/3 = 1
If Al DIDN'T get Bob's hat in the first step, then ONE of the remaining 3 hats is Bob's hat, which means P(Bob doesn't get his hat) = 2/3

The next step is even trickier.
What is P(Cal doesn't get his hat)?
This probability depends on whether Cal's hat was given to either Al or Bob in the first two step.
And so on....

NOTE: As I mention in my forum post, I created this question to highlight many students' tendency to avoid listing and counting as a possible approach. On test day, you will likely be able to use more than one approach (listing & counting, counting strategies, and probability rules), but you should always remember that it's possible that listing & counting MAY be the fastest approach.

Not specific to this exact video - But can you remind me the logic of using the slot method e.g. for https://gmatclub.com/forum/a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
I get that it is 8*6*4/3!,
we are divided by a 3! denominator because it is a combination problem/order does not matter, but what is the intuition behind 3 factorial? 3 decision steps?...
gmat-admin's picture

Question link: https://gmatclub.com/forum/a-committee-of-3-people-is-to-be-chosen-from-...

The 8*6*4/3! solution you're referring to is just one approach.

In this approach, we first choose ANY person (out of all 8 people)
This means the selected person's spouse cannot be chosen for this committee.

Next, we choose someone from the remaining 6 people.

Then we eliminate that person's spouse from the options and then choose someone from the remaining 4 people.

At this point, we need to check for duplication.

Notice that, if the 3 couples are AB CD EF GH, then the above approach allows for the following committee selections (in order of selection):

So, we've counted the A, C and E committee 6 times (Aside: we can arrange 3 objects in 3! ways and 3! = 6)
In fact, we've counted every possible committee 6 times.

To account for this duplication, we must divide by 3! (aka 6)

Here's my (different) approach: https://gmatclub.com/forum/a-committee-of-3-people-is-to-be-chosen-from-...


Hi Brent. I would like to know how the calculate the numerator for this problem - https://gmatclub.com/forum/if-3-different-integers-are-randomly-selected-from-the-integers-from-297381.html

Hi Brent,
Can you please share your solution to this question?



Hi Brent,
Is this an official gmat question? If yes then like you said that it is unlikely that on the test day 4 choices will get eliminated, then in such a case, how to proceed with this question?

gmat-admin's picture

Question link: https://gmatclub.com/forum/if-3-different-integers-are-randomly-selected...

Hi Kashaf,
This isn't an official question; it's one I created to show the utility of calculating the denominator first.

Some of the reasons to first calculate the denominator include:
1) If you're running behind on time, knowing the denominator might help eliminate some answer choices before making a guess and making up some lost time (if you're super lucky, you MIGHT eliminate 4 answer choices)

2) If you're unable to calculate the numerator (aside: calculating the numerator is often harder than calculating denominator), determining the denominator first may help eliminate some answer choices before making a guess.

For this question, calculating the numerator is quite tricky. Here's my full solution: https://gmatclub.com/forum/if-3-different-integers-are-randomly-selected...



Is there another method using which we can solve this question ?

gmat-admin's picture

Question link: https://gmatclub.com/forum/if-an-integer-n-is-to-be-chosen-at-random-fro...

Bunuel's solution (https://gmatclub.com/forum/if-an-integer-n-is-to-be-chosen-at-random-fro...) is a little bit different from mine, but I think most students would want to text out his assumptions just to be certain (in which case, the solution looks a lot like mine)


Hi Brent,

For me, the listing approach is difficult to use cause I am not to list down all scenarios correctly and in adequate time.

Could you share an approach to tackle this problem.
gmat-admin's picture

I know that many students find listing and counting to be a second-class strategy, but in many cases it's either the fastest approach or the ONLY approach. So it's important to work on those skills.

Also, in lots of cases (e.g., this question: https://www.gmatprepnow.com/module/gmat-counting/video/774), we can abandon listing and counting once we recognize a pattern.

My solution (https://gmatclub.com/forum/if-an-integer-n-is-to-be-chosen-at-random-fro...) involves listing and counting.

Although Bunuel's solution (https://gmatclub.com/forum/if-an-integer-n-is-to-be-chosen-at-random-fro...) is different, I think most students would test the assumptions in that solution before committing to an answer. So, even Bunuel's solution would likely involve some listing (and counting).

Hi Brent,

I guess I didn't structure my question correctly.

The listing approach works for me and sometimes its the only option.

I wanted to ask you how could I make myself better at it - so my question is - how can I quickly list down all the possible scenarios without missing out on any of them in the limited time we have.

I am asking this especially for the tough questions.
gmat-admin's picture

The trick is listing possible outcomes in a systematic way so that no outcomes are missed.

For example, let's say we totally forgot how to calculate combinations, and we must determine the total number of ways to create a 2-person committee from 5 employees.

Let's say the five employees are A, B, C, D and E.

One possible systematic approach is to always list prepared in alphabetical order.
Furthermore, we'll start by exhausting all possible outcomes featuring employee A.
We get:

Now we'll list all possible outcomes featuring employee B.
Since we've already listed AB, we'll start with BC and continue from there:

Then we get:

And finally:

To get more practice, go to the Listing and Counting video Lesson (https://www.gmatprepnow.com/module/gmat-counting/video/773), and try solving the linked questions in the Reinforcement Activities box by listing and counting.


Thanks Brent!

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