Lesson: Listing vs Counting vs Probability Rules

Comment on Listing vs Counting vs Probability Rules

you are helping me beyond measure. Much appreciation to your work :)

In the first method for the committee question, why do we use combination? In the given set of people the sequence is relevant in the sense that a male could have been selected. Then why 6C2?
gmat-admin's picture

For the numerator, we're trying to determine the total number of outcomes that satisfy the requirement that both committee members are women.

So, we need only consider the number of ways to select 2 women from 6 women (6C2)

A certain junior class has 1,000 students and a certain senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair ?
(A) 3/40,000
(B) 1/3,600
(C) 9/2,000
(D) 1/60
(E) 1/15
gmat-admin's picture

I have answered that question here: http://www.beatthegmat.com/probability-t106326.html

Please let me know if you have any questions.

Hi Brent,

For the Coin Q how would I setup the calculation if I wanted to use Counting? I am not referring to listing the possible outcomes but using the Combinations/Counting technique.

Thanks & Regards,
gmat-admin's picture

Here's one approach:

First, we'll use the complement to write:
P(at least 1 heads) = 1 - P(no heads)
= 1 - P(both tails)

Let's now calculate P(both tails)

P(both tails) = (# of outcomes with both tails)/(TOTAL # of possible outcomes)

DENOMINATOR: TOTAL # of possible outcomes
Stage 1: select outcome for 1st toss. Since we can get either heads or tails, this stage can be accomplished in 2 ways.
Stage 2: select outcome for 2nd toss. Since we can get either heads or tails, this stage can be accomplished in 2 ways.
So, TOTAL # of possible outcomes = (2)(2) = 4

NUMERATOR: # of outcomes with both tails
Stage 1: get tails on 1st toss: this can be accomplished in 1 way
Stage 2: get tails on 2nd toss: this can be accomplished in 1 way
So, # of outcomes with both tails = (1)(1) = 1

So, P(both tails) = 1/4

This means P(at least 1 heads) = 1 - 1/4 = 3/4


Thanks Brent!! I am assuming a typo in the paragraph beginning with NUMERATOR. I think you mean "get TAILS on 1st toss" and not HEADS. Please correct me if I am wrong.

gmat-admin's picture

Good catch, thanks!
I edited my response.


HI Brent, on the coin toss question, are you saying ORDER matters? T H is DIFFERENT from H T? and hence, counted as 2 separate outcomes? why cant we treat it as 1 outcome? meaning 1 H and 1 T, irrespective of order? thanks
gmat-admin's picture

Yes, I'm saying that the 1st toss is different from the 2nd toss.
If you try to treat the order as not mattering, you get a problem trying to count the number of possible outcomes (try it and you'll see)


Hello for the president question I did:

P(Harry is not president) AND (1 - P(Not Treasurer AND Not Secretary))

= 9/10 x (1- 8/9 x 7/8))
= 9/10 x (1 - 7/9)
= 9/10 x 2/9
= 2/10 = 1/5

It's probably more convoluted, but is the reasoning correct?
gmat-admin's picture

Question link: https://www.beatthegmat.com/president-t111434.html

That is a little convoluted, but the reasoning is sound - nice job!!


Hi Brent. Firstly, I'm finding the videos really helpful! I'm just wondering if/where you have a video on combinatorics and probability - I can't spot one. Thanks!
gmat-admin's picture

Combinatorics and Counting are the same thing :-)
Here's the counting module: https://www.gmatprepnow.com/module/gmat-counting
Here's the probability module: https://www.gmatprepnow.com/module/gmat-probability


Hi Brent,


In this question, I could figure out that we need 1 - (all three teas).

But after that I couldn't figure out which method to follow ( Counting, Combination or Probability formula)

Also when I counted the possible combinations for (all three teas).

There seemed only three options, is this thought correct?


Thank you
gmat-admin's picture

Question link: https://gmatclub.com/forum/at-a-blind-taste-competition-a-contestant-is-...

If we say that order does not matter, then there are three possible outcomes such that the contestant tries all three teas (XYZX, XYZY, and XYZZ).
If we go with this approach, then it looks like we're also treating the three cups containing the same tea as 3 IDENTICAL cups.
This becomes problematic very quickly when it comes to determining the total number of possible outcomes.

As far as how to solve this, I think there are two really nice methods already demonstrated:
1) Bunuel's solutions (at https://gmatclub.com/forum/at-a-blind-taste-competition-a-contestant-is-...) use combinations.
2) Cledgard's solution (athttps://gmatclub.com/forum/at-a-blind-taste-competition-a-contestant-is-...) uses probability rules.

I can't think of any solutions that are better than those ones.

Hi Brent,
Can you please provide solution to this question?

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