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## Comment on

Probability of Event A AND Event B## can we do this in a breaking

like for the denominator we have 7C2 and for the numerator we have first ball green times the second ball green 4x3=12

12/21

I don't know where I went wrong

thanks in advance

## This is a common error.

This is a common error.

You are treating the denominator as though order doesn't matter (7C2), but you are treating the numerator as though order does matter (4x3)

In the numerator, you have 4 x 3, which suggest there are 4 ways to select the first green ball and 3 ways to select the 2nd green ball. So, if we label the green balls A, B, C, and D, your calculations are essentially saying that selecting green ball A and then green ball B is DIFFERENT from selecting green ball B and then green ball A.

If we want to go the route were order matters, then your numerator (4x3) is correct, but we need to redo your denominator. If order matters, there are 7 ways to select the first ball and 6 ways to select the second ball. So, for the denominator, we get 7x6

So, P(2 green) = (4x3)/(7x6) = 12/42 = 2/7

## What do you think of this

P (G and G)= Possible 2 G balls to occur/Total possibilities

Calculating the Denominator: 7C2= 21

Calculating the Numerator: 4C2 (we have 4 green balls) = 6

So, P (G and G)= 6/21 = 2/7

Is this a correct method?

## Perfect approach.

Perfect approach.

Most probability questions can be solved using either counting methods or probability rules.

The video lesson is about applying probability rules to solve the question, but your solution is just as valid.

Great work!

## Hi Brent,

Could you please present a scenario in which the method suggested by Zoser(4C2/7C2) might actually fail? Intuitively this is the method that clicks for me while solving such questions and hence I was wondering if there is any scenario in which this method might fail and I might have to use P(A and B) formula.

Thanks & Regards,

Abhirup

## Tough question, Abhirup!

Tough question, Abhirup!

Counting methods (like the one demonstrated by Zoser) can often be used to solve probability questions.

These methods will work whenever they satisfy the given conditions.

So, for example, 4C2/7C2 works here because the order in which we select the balls does not matter. So, we can use combinations. If the order DID matter, then we wouldn't be able to use combinations.

It really depends on the context of the question.

Cheers,

Brent

## Hi Brent,

Please help me with this probability question:

If a number between 0 and 1\2 is selected at random, which of the following will the number most likely be between?

(A) 0 and 3/20

(B) 3/20 and 1/5

(C) 1/5 and 1/4

(D) 1/4 and 3/10

(E) 3/10 and 1/2

## We're looking for the biggest

We're looking for the biggest range of values.

To make it easier to spot the biggest range, let's rewrite all of the fractions with the same denominator.

(A) 0/20 and 3/20 RANGE = 3/20

(B) 3/20 and 4/20 RANGE = 1/20

(C) 4/20 and 5/20 RANGE = 1/20

(D) 5/20 and 6/20 RANGE = 1/20

(E) 6/20 and 10/20 RANGE = 4/20

Answer: E

## Hi Brent,

Even probability of selecting yellow ball first is 1/25, can you explain this. Not able to understand how can probability be same of selecting yellow ball first or second (when we select it second time we have only 24 balls, so this probability should be more)

## You're referring to the

You're referring to the question at https://www.gmatprepnow.com/module/gmat-probability/video/751

There are 25 balls and 1 of them is yellow. 2 balls are selected WITHOUT replacement.

We want P(2nd ball is yellow).

In order for the 2nd ball to be yellow, the 1st ball cannot be yellow.

So, P(2nd ball is yellow) = P(1st ball is NOT yellow AND 2nd ball IS yellow)

= P(1st ball is NOT yellow) x P(2nd ball IS yellow)

= 24/25 x 1/24

= 1/25

Does that help?

## How to solve it with the

## I believe you're referring to

I believe you're referring to the question at https://www.gmatprepnow.com/module/gmat-probability/video/751

There are 25 balls, and 1 of them is yellow.

2 balls are selected WITHOUT replacement.

We want P(2nd ball is yellow).

So, we can write:

P(2nd ball is yellow) = (# of outcomes where 2nd ball is yellow)/(total # of outcomes)

---------------------------------

DENOMINATOR: TOTAL # OF OUTCOMES

Stage 1: Choose ANY ball for 1st selection

There are 25 balls, so stage 1 can be completed in 25 ways

Stage 2: Choose ANY ball for 2nd selection

There are 24 balls remaining, so stage 2 can be completed in 24 ways

By the Fundamental Counting Principle (FCP), we can complete both stages in (25)(24) ways

----------------------------------

NUMERATOR: # OF OUTCOMES WHERE 2nd BALL IS YELLOW

Stage 1: Choose any non-yellow ball for 1st selection

There are 24 non-yellow balls, so stage 1 can be completed in 24 ways

Stage 2: Choose a yellow ball for 2nd selection

There is only 1 yellow ball, so stage 2 can be completed in 1 way

By the FCP, we can complete both stages in (24)(1) ways

------------------------------------

So, P(2nd ball is yellow) = (24)(1)/(25)(24)

= 1/25

Does that help?

Cheers,

Brent

## In determining the

Thanks a lot

## In this case, the order in

In this case, the order in which we select the balls matters. That is, selecting a blue ball first and selecting a yellow ball second is different from selecting a yellow ball first and selecting a blue ball second.

As such, we cannot use combinations.

Cheers,

Brent

## Did not understand this one

https://gmatclub.com/forum/two-children-one-from-each-grade-will-be-chosen-at-random-from-4-chi-239052.html

Can you explain like you would to a layman please?

## I'm happy to help.

I'm happy to help.

Question link: https://gmatclub.com/forum/two-children-one-from-each-grade-will-be-chos...

Did you see my solution at https://gmatclub.com/forum/two-children-one-from-each-grade-will-be-chos... ?

Can you review my solution, and let me know at points that need clarification.?

Cheers,

Brent

## I don't understand the answer

Two children, one from each grade, will be chosen at random from 4 children from the 5th grade class and 4 children from the 4th grade class to win a prize. Each class has 3 eligible girls and one eligible boy. What is the probability that one boy and one girl will be chosen?

Any help?

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/two-children-one-from-each-grade-will-be-chos...

Please let me know if you need any clarification.

Cheers,

Brent

## Hi Brent,

This is an easy one but I am just not getting the right answer. Source is Kaplan's test.

Attorneys are questioning 11 potential jurors about their ability to serve on a jury and decide a case fairly. Seven individuals will be selected to serve on the jury. If each potential juror has an equal chance of being selected, what is the probability that Tamara and Inga are chosen for the jury?

2/11

2/7

21/55

4/11

34/55

## Here's my full solution:

Here's my full solution: https://gmatclub.com/forum/attorneys-are-questioning-11-potential-jurors...

Cheers,

Brent

## will Dependent Probability be

thanks

## On test day, you'll see

On test day, you'll see questions in which the events are dependent, and you'll see questions in which the events are independent.

## Hi Brent, I would appreciate

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

1) I understand that we could solve 1) as: n/10 * (n-1)/9. I am just confused, why cant we do Nc2/10c2. Meaning why cant we combine the combinations as Nc2/10c2?

2) I am confused why we cant treat part 2) as a pizza topping qs where order is NOT important. Meaning an onion picked first and bell pepper picked second is the same as bell pepper picked first and onion picked second. Meaning the fact is that there are 2 toppings on the pizza. Similarly, why cant we just say that one is non defective and one is defective. Why do we have to add the 2 probabilities that first is defective and second non defective + second is defective and first is non defective

Thanks

## Question link: https:/

Question link: https://gmatclub.com/forum/a-box-contains-10-light-bulbs-fewer-than-half...

Great questions!

(1) Your suggested approach also works, since (n/10)[(n-1)/9] = nc2/10c2

If you perform the calculations, you'll find they are equal.

(2) We can use either approach (order matters or order does not matter), just as long as treat the numerator and denominator the same way.

If we say that order does NOT matter, then our calculations are as follows:

There are n defective bulbs and 10-n good bulbs

So, we can select a defective bulb in n ways and a good bulb in 10-n ways

Total number of way to get 1 bad bulb and 1 good bulb = (n)(10-n)

For the DENOMINATOR, we can select 2 bulbs from 10 bulbs in 10C2 ways (= 45 ways)

So, we get: (n)(10-n)/45 = 7/15

Solve to get: n = 7 OR n = 3

However, since we're told FEWER THAN HALF are defective, we know that n = 3

Does that help?

Cheers,

Brent

## Hi Brent, could u pls help

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

I am able to get 10 * 8 * 6= 480. However, why do I have to divide this by 6? I dont understand.

I also understand ORDER is NOT important in such a qs. Why is that? Cant we pick the first person and second person and third person in DIFFERENT ways? When is ORDER important and when is it NOT important? I am confused. Pls guide. Thanks

## Choosing Person A, then

Choosing Person A, then Person B then Person C to be on the committee yields the same outcome as choosing Person B, then Person C then Person A to be on the committee.

In both cases, A, B and C are on the committee. As such, we must treat the question as though order does not matter.

To get a better idea of what I'm saying, consider an easier question.

Let's say there are no couples. Instead, we just have 3 people: A, B and C

In how many different ways can we create a 3-person committee?

If we follow your approach, we get 3x2x1 = 6

Of course, there is only 1 way to create a 3-person committee

Does that help?

Here's my full solution to the question: https://gmatclub.com/forum/if-a-committee-of-3-people-is-to-be-selected-...

Cheers,

Brent

## Hi Brent,

https://gmatclub.com/forum/a-box-contains-10-light-bulbs-fewer-than-half-of-which-are-defective-99940.html

My doubt is related to the language of the question. The question clearly mentions that both the bulbs are taken out simultaneously. Now, my question is whether this should have any effect on the question as opposed to "2 bulbs are taken out one by one, without putting the first one back"?

Should n/10*(n-1)/9 still hold true?

Thanks, you're always a great help!

## Question link: https:/

Question link: https://gmatclub.com/forum/a-box-contains-10-light-bulbs-fewer-than-half...

Selecting 2 bulbs simultaneously, and selecting the 2 bulbs one at a time without replacement will yield the same results.

Cheers,

Brent

## Hello there again Brent.

I understand how to solve this, but tried a different approach of trying to get the complement and deducing from 1. Why doesn't it work.

Example: Chance of choosing two defective pens - 3/12 x 2/11 and deducting this from 1. It gives us 21/22. Why doesn't this equate to the 6/11? Just curious.

https://gmatclub.com/forum/in-a-box-of-12-pens-a-total-of-3-are-defective-if-a-customer-buys-207395.html

## Question link: https:/

Question link: https://gmatclub.com/forum/in-a-box-of-12-pens-a-total-of-3-are-defectiv...

You're correct to say that: P(selecting two defective pens) = (3/12)(2/11) = 1/22

This means P(NOT selecting two defective pens) = 1 - 1/22 = 21/22

However, "NOT selecting two defective pens" isn't the same as "selecting zero defective pens"

Here's why:

If we select two pens at random, there are three possible outcomes:

case 1) We select 2 defective pens

case 2) We select 1 defective pen and 1 working pen

case 3) We select 0 defective pens

So, "NOT selecting two defective pens" is the same as case 1 NOT occurring

If case 1 does NOT occur, then EITHER case 2 occurred OR case 3 occurred.

In other words, "NOT selecting two defective pens" = "selecting 1 defective pen and 1 working pen OR selecting two working pens"

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