# Lesson: Rewriting Questions

## Comment on Rewriting Questions

### Hi, can you pls explain how

Hi, can you pls explain how did you get the answer 3/7 to the qts. u asked to solve by ourselves. I got an answer 25/7. Thank you ### Sure thing.

Sure thing.
P(product is positive) = P(both #s are positive) + P(both #s are negative)

P(both #s are positive) = P(1st # is positive) x P(2nd # is positive| 1st # is positive)
= 3/7 x 2/6
= 1/7

P(both #s are negative) = P(1st # is negative) x P(2nd # is negative| 1st # is negative)
= 4/7 x 3/6
= 2/7

So, P(product is positive) = 1/7 + 2/7
= 3/7 ### By the way, ALL probabilities

By the way, ALL probabilities must be greater than or equal to zero and smaller than or equal to 1.
The probability that you found (25/7) is outside this range.

### Hi, Can you please explain

Hi, Can you please explain the solution to 4 Boys and 4 Girls question if we do not select complement method and go via OR probability ### Sure thing. We need to

Sure thing. We need to consider three possible outcomes:
1) 1st child selected is a boy, and 2nd child selected is a girl.
2) 1st child selected is a girl, and 2nd child selected is a boy.
3) 1st child selected is a boy, and 2nd child selected is a boy.

So, P(at least 1 boy) = P(1st boy and 2nd girl OR 1st girl and 2nd boy OR 1st boy and 2nd boy)
= P(1st boy and 2nd girl) + P(1st girl and 2nd boy) + P(1st boy and 2nd boy)
= (4/8)(4/7) + (4/8)(4/7) + (4/8)(3/7)
= 16/56 + 16/56 + 12/56
= 44/56
= 11/14

### Hi Brent! Why the events

Hi Brent! Why the events inside the big OR probability formula are mutually exclusive? ### There are three possible

There are three possible events/outcomes that satisfy the condition of getting at least one boy:
1) 1st child selected is a boy, and 2nd child selected is a girl.
2) 1st child selected is a girl, and 2nd child selected is a boy.
3) 1st child selected is a boy, and 2nd child selected is a boy.

Two events are mutually exclusive if both events cannot occur at the SAME TIME.

If we select 2 children, it is impossible for events 1 and 2 (above) to occur at the same time. Likewise, it's impossible for events 1 and 3 (above) to occur at the same time. Etc.

This means the 3 events are mutually exclusive.

### If we use complement then it

If we use complement then it would be 4/8 * 3/7 = 3/14
So 1 - 3/14 = 14/14 - 3/14 = 11/14 Perfect!

### I used complement, at least

I used complement, at least one number must be negative to get negative product and probability of getting at least one negative number is 4/7, using complement formula we get 1-4/7=3/7 i.e. both number will be positive or negative. Hope this is okay? ### That's a perfectly valid

That's a perfectly valid approach.

### Hi Brent,

Hi Brent,

How to solve the question with 4 boys and 4 girls using counting techniques?

Thank you.

Thank you. ### You're referring to the

You're referring to the question that starts at 3:05

I'd still use the complement.

That is, P(at least 1 boy) = 1 - P(both girls)

P(both girls) = (# of ways to select 2 girls from 4 girls)/(# of ways to select 2 children from 8 children)
= (4C2)/(8C2)
= 6/28
= 3/14

So, P(at least 1 boy) = 1 - P(both girls)
= 1 - 3/14
= 11/14

### I solved 4 boys and 4 girls

I solved 4 boys and 4 girls problem using combination, as you explained above. But want to know can we solve all problems either using combination or probability formulas (without combination nCk) ### In many cases, we can treat

In many cases, we can treat the counting/probability question either way.

If we say that order does NOT matter (as I have done in the video solution), then we must calculate both the numerator and denominator as though order does NOT matter.

If we say that order matters, then we must calculate both the numerator and denominator as though order matters.

Try it with this question and see how you do.

### In a garden, there are three

In a garden, there are three blue flowers, three red flowers, three green flowers, and three pink flowers. What is the probability that a florist will choose three flowers of the same color when randomly picking three flowers?

Hi Brent - I saw your solution for this questions and it was 1 x 2/11 x 1/10 = 1/55 but it is multiplied by 1 and not by 1/12 x 2/11 x 1/10 - Could you explain this please? ### You bet.

You bet.

The key here is to recognize what needs to happen with each random selection of flowers.

So, for example, if we want all 3 flowers to be the same color, what MUST happen on the 1st selection?

Well, the first selection is inconsequential. That is, we don't NEED the first selection to be any specific color.

For example, if the 1st flower selected were blue, then that would be fine. We'd just need the 2nd and 3rd selected flowers to be blue as well (in order for all three flowers to be the same color).

Likewise, if the 1st flower selected were red, then that would be fine too. We'd just need the 2nd and 3rd selected flowers to be red as well (in order for all three flowers to be the same color).

And so on.

So, we get: P(all the same color) = P(1st flower is ANY color AND 2nd flower matches 1st flower AND 3rd flower matches 1st flower)
= P(1st flower is ANY color) x P(2nd flower matches 1st flower) x P(3rd flower matches 1st flower)
= (1)(2/11)1/10)
= 1/55

P(1st flower is ANY color) = 1, because the first flower can be ANY color

P(2nd flower matches 1st flower) = 2/11, because once the first flower has been selected, there are 11 flowers remaining, and only 2 of those 11 flowers are the same color as the first flower selected.

P(3rd flower matches 1st flower) = 1/10, because once the first and second flowers have been selected, there are 10 flowers remaining, and only 1 of those 10 flowers are the same color as the first flower selected.

Here's my full solution: https://gmatclub.com/forum/in-a-garden-there-are-three-blue-flowers-thre...

### Hey Brent:

Hey Brent:
Is it also possible to solve the last question (the one with the students) like this:

I also choose to solve it with the complement strategie. I first calculated the denominator: 2 Students can be choosen out of 8 Students in 8C2 ways. (=28)

Then I calculated the numerator "In how many ways can 2 girls be chosen out of 4 girls"? (=4C2 = 6)

This leaves us with 6/28, The compliment is 22/28 = 11/14 ### Hi Alicia,

Hi Alicia,

You're referring to the question that begins at 3:01 in the video.

Your counting approach is perfectly valid. Nice work!

Cheers,
Brent

Hey Brent,

Would you mind please explaining why we don't multiply 2 to the equation such that it is 2 * 30/500 * 1/800? My logic was that you can either select the first sibling from the business school or the law school. Would be awesome if you had a way of telling when we need to account for different cases, and why that doesn't apply here

Thank you! The two cases you are describing are identical, so we can't count them twice.

To illustrate what I mean, consider a modified question where the business school has only 1 student (Joe), and the law school has 1 student (Jane). Joe and Jane are siblings.

As you can see, the probability is 1 that two siblings are selected. However, if we treat your two cases as being different, the probability becomes 2.

Does that help?

Cheers,
Brent

### The way I am thinking about

The way I am thinking about it also is that it seems like the business student and law student picked are coming from different sets? As we have seen earlier in your lessons, this mitigates the need to account for other cases (contrary to what we do in questions for eg where we have 3R 3B flowers and have to find the probability of getting 2 flowers of the same colour with random pickings) ### It really comes down to begin

It really comes down to begin how we write our probability equations to match the given conditions. If we create an OR probability, we need to ensure that we're not considering the same outcome more than once.

In this question, selecting Joe first and Jane second is considered the exact same outcome as selecting Jane first and Joe second. In fact we could also select Joe and Jane simultaneously, but that outcome would be considered the same as the other two outcomes.

If you could provide a specific example of a question where we handle things differently, we can go from there.

Cheers,
Brent

### https://www.gmatprepnow.com

https://www.gmatprepnow.com/module/gmat-probability/video/762

The Q in the video above is an example of how we go through each of the three colours in the OR setup. Had there been an equal number of balls for each colour, we would have multiplied the number of colours with the P(2 balls drawn having the same colour)

I had applied this logic to the business / law school question initially. This was wrong because we are picking a predetermined Jane - Joe combo from the two sets. So whether we pick Jane from the business school (30/500) and then Joe (Jane's pair; 1/800) from the law school, it would be the same as picking Joe first and then Joe's pair - Jane second. Thus the order doesnt matter.

Please let me know if I am understanding the difference correctly.

Thank you,
Neel ### In that video question (at

In that video question (at https://www.gmatprepnow.com/module/gmat-probability/video/762), we set up the equation as follows:

P(both balls are same color) = P(the balls are both green OR the balls are both red OR the balls are both yellow)

Here, the three events (both green, both red and both yellow) are different events. Since they are different events, we have no duplication. So, everything is fine.

"Had there been an equal number of balls for each colour, we would have multiplied the number of colours with the P(2 balls drawn having the same colour)"

That is true, but the different events (both red, both yellow, both green etc are still different events)

In your solution to the business/law school question, the events aren't different. As you say, the order doesn't matter here.

Cheers,
Brent

### the first example could have

the first example could have been more easily solved using combinations. The overall possibilities are 7 choose 2. Both being negative are 4 choose 2 and both being positive are 3 choose 2. 4 choose 2 plus 3 choose 2 = 9. 7 choose 2 = 21. 9/21 = 3/7 ### You're absolutely right.

You're absolutely right.

That said, the purpose of the video lesson is to learn how to rewrite questions so that we can apply the various probability formulas.

Cheers,
Brent

### Hi Brent, would you mind

Hi Brent, would you mind offering me some insight as to when we would/could use the "nCr" (AKA combination formula) for probability questions?

I can't seem to figure out what scenarios this formula would apply to with regards to probability Qs. If you don't mind providing 2-3 generic examples, I would appreciate it!

Thanks as always :). ### Hey brownpure,

Hey brownpure,

Here are a couple of examples:
- https://www.beatthegmat.com/8-volunteers-for-charity-event-t276838.html

Cheers,
Brent ### Thank you Brent, but I

Thank you Brent, but I believe you may have misinterpreted my question (That is my fault!).

What I mean is: When/what is an appropriate scenario to use the Combination formula FOR probability questions?

So from my understanding, one example could be to find the probability of selecting a specific member for a committee?

Thank you! ### Tough question!

Tough question!

Many probability questions can be solved using either counting techniques (e.g., combinations) or probability rules. The hard part is determining those situations that are best suited for counting techniques and which one are best suited for probability rules..

That said, for questions involving committees (or other questions in which the order in which we make selections does not matter), it's LIKELY that we can use combinations.

Does that help?

Cheers,
Brent

### Yes it does, thanks as always

Yes it does, thanks as always!

### Hi Brent, having trouble

Hi Brent, having trouble understanding why my solution isn't correct for the below question.

https://gmatclub.com/forum/if-3-different-numbers-are-selected-from-the-first-8-prime-numbers-wh-277804.html

My Solution: Even + odd + odd = even

P(even)*P(odd)*P(odd) = P(1 way of selecting 2)*P(7/8)*P(6/7) = 3/4

But if I try to solve it using complement theory I get 3/8. You're on the right track, but there are a few things missing.

P(1st # is even AND 2nd # is odd AND 3rd number is odd)
= P(1st # is even) x P(2nd # is odd) x P(3rd number is odd)
= 1/8 x 1 x 1
= 1/8

A few things to note:
P(1st # is even) = 1/8, because there are 8 numbers to choose from, and only 1 of them is even

P(2nd # is odd) = 1, because once we select the 2 as the 1st number, all of the remaining 7 numbers are odd. So, it's GUARANTEED that the 2nd number (given that the 1st selected number is 2).

IMPORTANT: At this point, we have examined only 1 way to get one even number and two odd numbers. That is: 1st # is even AND 2nd # is odd AND 3rd number is odd

There are two more ways to get one even number and two odd numbers. They are:
- 1st # is odd AND 2nd # is even AND 3rd number is odd
- 1st # is odd AND 2nd # is odd AND 3rd number is even

When we calculate these two probabilities, we get:
- P(1st # is odd AND 2nd # is even AND 3rd number is odd) = 1/8
- P(1st # is odd AND 2nd # is odd AND 3rd number is even) = 1/8

So, P(sum is even) = 1/8 + 1/8 + 1/8 = 3/8

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

in the below question, my answer is coming to 11/10;

Matched pair = P(R1 and R2) and P(R2 and R1) or P(D1 and D2) and P(D1 and D3) and P(D2 and D3). I took this step following the above solution to the question of selecting 3 numbers. Can you explain why no. of ways to select doves and rabbit is not important here v/s the above number question.

https://gmatclub.com/forum/a-magician-has-five-animals-in-his-magic-hat-3-doves-and-140229.html ### Let's keep going with your

Let's keep going with your solution:

P(Matched pair) = P(R1 and R2) OR P(R2 and R1) OR P(D1 and D2) OR P(D1 and D3) OR P(D2 and D3)

NOTE: We're missing a few at the end.
It should be....

P(Matched pair) = P(R1 and R2) OR P(R2 and R1) OR P(D1 and D2) OR P(D1 and D3) OR P(D2 and D3) OR P(D2 and D1) OR P(D3 and D1) OR P(D3 and D2)
= P(R1 and R2) + P(R2 and R1) + P(D1 and D2) + P(D1 and D3) + P(D2 and D3) + P(D2 and D1) + P(D3 and D1) + P(D3 and D2)
= (1/5)(1/4) + (1/5)(1/4) + (1/5)(1/4) + (1/5)(1/4) + (1/5)(1/4) + (1/5)(1/4) + (1/5)(1/4) + (1/5)(1/4)
= 1/20 + 1/20 + 1/20 + 1/20 + 1/20 + 1/20 + 1/20 + 1/20
= 8/20
= 2/5
WORKS!!!

### Hi Brent,

Hi Brent,

In the below questions, statements says 'atleast a 5 or a 1 on either die', doesn't that mean that to win the user needs a 5, 6 or 1 on either die? https://gmatclub.com/forum/in-a-game-one-player-throws-two-fair-six-sided-die-at-the-151956.html I'm not sure where the 6 came from.
The question reads "If the player receives at least a five or a one on EITHER die, that player wins"

So, here are a few outcomes in the form (1st die, 2nd die) that would be a winning roll: (1, 4), (2, 5), (3, 1), (6, 1), (5, 5), (2, 1), etc.

Does that help?

Cheers,
Brent

### Oh shoot, I misinterpreted

Oh shoot, I misinterpreted the statement. I thought it says x>=5 or x=1 in either die to win. ### Ahhhh, I was wondering where

Ahhhh, I was wondering where the 6 came from :-)

### Hi Brent

Hi Brent

Is there a quick way or formula to figure out this question: If you roll two fair six-sided dice, what is the probability that the sum is 4 or higher?

Thanks ### Notice that P(sum is 4 or

Notice that P(sum is 4 or more) = P(sum = 4 or 5 or 6 or 7 or ....or 11 or 12)
That's a LOT of different probabilities we'll need to find.

However, P(sum is 3 or less) = P(sum = 2 or 3)

So, the fastest route is to use the complement.
That is: P(sum is 4 or more) = 1 - P(sum is NOT 4 or more)
In other words: P(sum is 4 or more) = 1 - P(sum is LESS THAN)
In other words: P(sum is 4 or more) = 1 - P(sum is 2 OR sum is 3)

So, all we have to do now is calculate P(sum is 2 OR sum is 3)

P(sum is 2 OR sum is 3) = P(sum is 2) + P(sum is 3)

I'll leave it to you to finish the calculations.
(the answer to the original question is 11/12)

Cheers,
Brent

### Thanks Brent!

Thanks Brent!

I was just looking for a quick formula or method of sorts (if there was a shortcut or another way.)so I didn't have to create a 6x6 grid. Just thinking how time consuming and stressing that would be under the pressure of a real test. ### The good thing about using

The good thing about using the complement is that you shouldn't need a 6x6 grid, since there are only 3 ways to get a sum of 3 or less:
(1,1)
(1,2)
(2,1)

Cheers,
Brent

### Hi Brent, would you mind

Hi Brent, would you mind explaining again the difference between dependent and independent events please? Thanks very much x ### Whether or not two events are

Whether or not two events are dependent or independent relies on whether the occurrence of one event affects the occurrence of the other event.

For example, let's say that a bag contains 3 black marbles and 1 red marble. Let's say I randomly remove one marble and then a second marble without replacement. Now let's examine two probabilities related to this experiment:
P(1st ball is red)
P(2nd ball is red)

What is P(2nd ball is red)?
Well, we know that after the first ball is drawn there are three balls remaining, but how many of those three balls are red?
The answer to that question DEPENDS on whether the first ball is red or black.
If the first ball is red, then P(2nd ball is red) = 0/3
If the first ball is black, then P(2nd ball is red) = 1/3
So, in this case the two events "1st ball is red" and "2nd ball is red" are DEPENDENT.

--------------------------
Here's another example:

Let's say a new experiment involves rolling a die and flipping a coin.
Now let's examine two probabilities related to this experiment:
P(the die displays 5)