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## Comment on

Listing vs Counting vs Probability Rules## you are helping me beyond

## In the first method for the

## For the numerator, we're

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

(A) 3/40,000

(B) 1/3,600

(C) 9/2,000

(D) 1/60

(E) 1/15

## I have answered that question

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,

Abhirup

## Here's one approach:

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

Cheers,

Brent

## Thanks Brent!! I am assuming

Regards,

Abhirup

## Good catch, thanks!

Good catch, thanks!

I edited my response.

Cheers,

Brent

## HI Brent, on the coin toss

## Yes, I'm saying that the 1st

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)

Cheers,

Brent

## Hello for the president

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?

## Question link: https://www

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

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

Cheers,

Brent

## Hi Brent. Firstly, I'm

## Combinatorics and Counting

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

Cheers,

Brent

## Hi Brent,

https://gmatclub.com/forum/at-a-blind-taste-competition-a-contestant-is-offered-3-cups-of-each-of-86830.html

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?

XYZX

XYZY

XYZZ

Thank you

## Question link: https:/

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?

https://gmatclub.com/forum/an-exam-consists-of-8-true-false-questions-brian-forgets-to-174966.html

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/an-exam-consists-of-8-true-false-questions-br...

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