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## Comment on

Introduction to Probability## in the set of {3,5,9,12,15,18

## Oh my bad. 5 is a prime

## For the linked question: http

Given that any multiple of 8 will also be a multiple of 2,

why can't we just divide 1000/8 to find the number of multiples of 8, and make that (125) the numerator over the denominator of 1000?

## Yes, we can. In fact that's

Yes, we can. In fact that's what I did in my solution.

However, my solution also includes a word of caution.

Notice what happens if the question were: If one number is chosen at random from the first 10 positive integers, then what is the probability that the number is a multiple of 2 and 8?

In this case, dividing 10 by 8 will not yield the correct answer. So, we need to be careful to not try to apply a generic approach to all similar questions.

## With the last practice

## If we were randomly selecting

If we were randomly selecting ONE person from the group of five people, then there would be 5 ways to do that:

1) A

2) B

3) C

4) D

5) E

We are randomly selecting TWO people from the group of five people. There are 10 ways to do that:

1) AB

2) AC

3) AD

4) AE

5) BC

6) BD

7) BE

8) CD

9) CE

10) DE

Another way to make this calculation is to use combinations, since the order in which we select the 2 people does not matter. For example, selecting A then B (AB) is the same as selecting B then A (BA)

More here: https://www.greenlighttestprep.com/module/gre-counting/video/787

## https://magoosh.com/gre/2012

The solution for the problem is not mentioned, or at least I cant view it. My understanding is to build a 90 degree triangle inside the circle with legs 2 each and then the hypotenuse is going to be 2 root 2. After that my mind goes numb. Can you please explain this.

## Question link: https:/

Question link: https://magoosh.com/gre/2012/gre-probability-practice-question-of-the-we...

Here's my full solution: https://greprepclub.com/forum/if-points-a-and-b-are-randomly-placed-on-t...

Cheers,

Brent

## Tossing a fair coin, the

## Given that there aren't many

Given that there aren't many possible outcomes when we flip 3 coins, a relatively fast and painless solution is to list all possible outcomes.

We'll list the outcomes in terms of H's and T's, where the 1st letter represents the 1st coin, the 2nd letter represents the 2nd coin, etc.

The outcomes are as follows:

1) HHH

2) HHT

3) HTH

4) THH

5) TTH

6) THT

7) HTT

8) TTT

Now check to see which outcomes have at least 2 heads.

They are outcomes 1, 2, 3 and 4

So, 4 of the 8 outcomes have at least 2 heads

P(at least 2 heads) = 4/8 = 1/2

Answer 1/2

Cheers,

Brent

## Hi Brent, for the example

I looked at it this way:

There are 5 people to choose from, so my denominator is 5-----> (some number, aka numerator/5)

The chances of Amir being chosen the first time around out of 5 people is one time, therefore 1/5.

The chances of Amir being chosen again a second time, once again, 1/5.

Add up both fractions 1/5 + 1/5 =2/5, or .40

Is this a valid approach?\

Thanks much!

## GREAT QUESTION! TOUGH

GREAT QUESTION! TOUGH QUESTION!

I read your post last night and, since your strategy didn't "feel" right to me, I started looking for counter-examples (using the same strategy you describe) to show the flaw in your strategy . . . but your strategy worked every time!

So, I decided to sleep on it.

Now that I've had plenty of time to think about it, I can tell you that the main issue with your solution is that it doesn't adhere to the general probability formula.

That is, P(Event A occurs) = (# of ways event A can occur)/(total number of possible outcomes)

For example, in your solution, you say "There are 5 people to choose from, so the denominator is 5"

However, what does the 5 represent?

It doesn't represent the total number of possible outcomes, since there are 10 ways in which we can choose TWO people from 5 people.

The same can be said of the numbers you're using in the numerator; they don't represent values needed to apply the general probability formula.

HOWEVER (and it's a big HOWEVER), even though the numbers you're using don't represent values needed to apply the general probability formula, those values do, indeed, work.

In fact, we can PROVE they work.

Based on your strategy, we can write: IF JOE IS AMONG N PEOPLE, AND 2 PEOPLE ARE RANDOMLY SELECTED, THEN P(JOE IS SELECTED) = 2/N

Here's the proof:

If there are N people, then we can select 2 people in NC2 (N choose 2) different ways.

NC2 = (N)(N-1)/2

So, this is our denominator

In how many of the (N)(N-1)/2 possible outcomes is Joe chosen?

Well, once we make Joe one of the selected people, there are N-1 people remaining to join Joe.

So, there are N-1 outcomes in which Joe is one of the selected people.

This is our numerator

So, P(Joe is selected) = (N-1)/[(N)(N-1)/2] = (N-1)[2/(N)(N-1)] = 2/N

Voila!!

We can even extend this property to say:

IF JOE IS AMONG N PEOPLE, AND 3 PEOPLE ARE RANDOMLY SELECTED, THEN P(JOE IS SELECTED) = 3/N

We can also say:

IF JOE IS AMONG N PEOPLE, AND 4 PEOPLE ARE RANDOMLY SELECTED, THEN P(JOE IS SELECTED) = 4/N

In general, we can say:

IF JOE IS AMONG N PEOPLE, AND K PEOPLE ARE RANDOMLY SELECTED, THEN P(JOE IS SELECTED) = K/N

Cheers,

Brent

## Hi Brent,

Could you explain more the last two lines on your solution of question number 11 from Reinforcement Activities

## Hi Omer,

Hi Omer,

In the future, please include the link of the question so I'm certain which question you're referring to.

My previous solution needed a little work in clarifying matters.

I have added a few more lines at the ends to help.

Please let me know what you think: https://greprepclub.com/forum/if-an-integer-greater-than-100-and-less-th...

Cheers,

Brent

## Thanks a million, dear Brent.

## Hi Brent ,

For https://greprepclub.com/forum/if-3-different-integers-are-randomly-selected-from-the-integ-13816.html

I didnt get this line

"When we check the answer choices, we see that only one answer choice (C) has a denominator that's a FACTOR of 440."

could you please help me how did you approach to option c

Thanks

## Question link: https:/

Question link: https://greprepclub.com/forum/if-3-different-integers-are-randomly-selec...

When we have a fraction that can be reduced to a simpler fraction, the denominator of the simplified fraction will always be a factor of the original fraction.

To see what I mean, let's simplify some fractions.

3/6 = 1/2. Notice that 2 is a factor of 6

15/20 = 3/4. Notice that 4 is a factor of 10

28/100 = 7/25. Notice that 25 is a factor of 100

12/440 = 3/110. Notice that 110 is a factor of 440

77/440 = 7/40. Notice that 40 is a factor of 440

So, if we simplify x/440, we know that the denominator of the simplified fraction must be a factor of 440.

Does that help?

## Hi Brent,

for this question: https://greprepclub.com/forum/one-person-is-to-be-selected-at-random-from-a-group-of-25-pe-8094.html

Isn't the number of males are 11 out of 25, as such should't be the number of males that are born before 1960 be: 0.28 x 11 (instead 0.28 x 25, since we know that among the total of 25, there are a non-male people).

## Question link: https:/

Question link: https://greprepclub.com/forum/one-person-is-to-be-selected-at-random-fro...

Great question!

It all comes down to the wording.

That is, it all comes down to the group of people from which we are selecting one person.

The question tells us that "the probability that the selected person will be a male who was born before 1960 is 0.28."

We're selecting 1 person out of all 25 people (not just 1 person from the 11 males)

So, if B equals the number of males born before 1960, then B/25 = 0.28

If the question had told us that we're randomly selecting one of the males, and that the probability that the selected person will be a male who was born before 1960 is 0.28, then we'd write: B/11 = 0.28

Does that help?

Cheers,

Brent

## Thank you Brent, yes it does!

## Hi Brent! Question on this

Why can't we use the p(A and B) = p(A) X p(B) rule for this?

P(divisible by 2) is 1/2 while P(divisible by 8) is 1/8, can we multiply the two to get 1/16th as the answer?

## Question link: https:/

Question link: https://greprepclub.com/forum/gre-math-challenge-15-if-one-number-is-cho...

The question: If one number is chosen at random from the first 1000 positive integers,then what is the probability that the number is a multiple of 2 and 8?

We can definitely use probability rules to solve this question (for students just starting probability, this particular concept is covered in the following video: https://www.greenlighttestprep.com/module/gre-probability/video/752)

p(A and B) = p(A) X p(B) when the two events are independent. In this case the two events are not independent.

That said we can always use the formula p(A and B) = p(A) X p(B|A)

So, P(# is a multiple of 2 AND 8) = P(# is multiple of 2) x P(# is multiple of 8 | # is multiple of 2)

= 1/2 x 1/4

= 1/8

Let's take a closer look at P(# is multiple of 8 | # is multiple of 2)

In other words: P(# is multiple of 8 GIVEN THAT # is multiple of 2)

If the number is a multiple of 2, then the number can be 2, 4, 6, 8, 10, 12, 14, 16, 18,...998, 1000

So, GIVEN THAT the number is in the set 2, 4, 6, 8, 10, 12, 14, 16, 18,...998, 1000, what is the probability that the number is divisible by 8?

Well, as you can see, 1 out of every 4 numbers is divisible by 8.

So, P(# is multiple of 8 | # is multiple of 2) = 1/4

Does that help?

## For this question: In a set

## There are 4 different

There are 4 different outcomes, but they aren't the same as those you have listed.

We can see why the outcomes are different if we label our sets as:

Set A: {1,2,4,6,7}

Set B: {2,3,4,5,6,7,8,9}

There are 4 different ways to get a sum of 11:

#1) select 2 from Set A and select 9 from set B

#2) select 4 from Set A and select 7 from set B

#3) select 6 from Set A and select 5 from set B

#4) select 7 from Set A and select 4 from set B

Notice that outcome #2 is different from outcome #4, since we're selecting the 4 and 7 from different sets each time.

## So it would be a mistake in

## Those two outcomes are

Those two outcomes are considered the same. In both cases, we're selecting a 4 from Set A and a 7 from Set B.

## https://greprepclub.com/forum

For this problem isnt 2,3 a different set than 3&2 so wouldnt we count 2&3,3&2, 2&5,5&2?

## Question link: https:/

Question link: https://greprepclub.com/forum/what-is-the-probability-that-the-sum-of-tw...

We have two options for this question:

1) treat the numerator and denominator as though order matters (e.g., 3 + 2 and 2 + 3 are different)

2) treat the numerator and denominator as though order does NOT matter (e.g., 5 + 3 and 3 + 5 are the same)

Either way, we'll reach the same answer.

1) Order matters:

There are 12 different outcomes: 2+3, 3+2, 2+5, 5+2, 2+7, 7+2, 3+5, 5+3, 3+7, 7+3, 5+7, 7+5

There are 8 non-prime sums : 2+7, 7+2, 3+5, 5+3, 3+7, 7+3, 5+7, 7+5

So, P(sum is not prime) = 8/12 = 2/3

2) Order doesn't matter:

There are 6 different outcomes: 2+3, 2+5, 2+7, 3+5, 3+7, 5+7

There are 4 non-prime sums : 2+7, 3+5, 3+7, 5+7

So, P(sum is not prime) = 4/6 = 2/3

The only time we get a problem is when we treat the numerator and denominator different (e.g., order matters for the numerator, and order doesn't matter for the denominator.)

## https://greprepclub.com/forum

For this problem if its a multiple of 2 and a multiple of 8 isnt it a multiple of 16? an couldnt we find the multiples of 16 starting from 16 ...992. All multiples of 16 are multiples of 2 & 8 right?

## Question link: https:/

Question link: https://greprepclub.com/forum/gre-math-challenge-15-if-one-number-is-cho...

Q: If its a multiple of 2 and a multiple of 8, isn't it a multiple of 16?

A: No. 8, 24, and 40 (etc) are multiples of 2 and 8, but they aren't multiples of 16

Since all multiples of 8 are also multiples of 2, we're really just looking for multiples of 8

## In your video in the last

## When using counting

When using counting techniques for probability questions, it's crucial that you use the same strategy for both numerator and denominator.

So, what does question, we can calculate the denominator using combinations. That is, we can select 2 people from 5 people in 5C2 ways (10 ways). Since we used combinations to calculate the denominator, we are assuming that order does not matter.

So, when calculating the numerator, we must assume that order does not matter.

In how many ways can we select two people so that Amir is one of those people?

There are 4 possible selections: AB, AC, AD, and AE.

So the probability = 4/10 = 0.4

Conversely, if we assume that order DOES matter, then there are 20 different ways to select two people from five people (5 x 4 = 20)

This means we must treat the numerator in the same way.

If order matters then there are 8 ways to select two people so that Amir is one of those people (AB, AC, AD, AE, BA, CA, DA, and EA.

So the probability = 8/20 = 0.4