# Lesson: The Complement

## Comment on The Complement

### Hello Brent,

Hello Brent,
If we only need 1 number to be even to have an even product, why is 1/2 an icorrect answer, since a pick of the "2" from the first set is sufficient to make the product even ? ### You are correct in saying

You are referring to the question that starts at 2:30.

You're correct in saying that P(selecting the "2" from the first set) = 1/2.
However, that is only one way to get an even product.

We can also get an even product by selecting the 3 from the first set and selecting the 8, 12, 18 or 20 from the second set.

A better way to illustrate the issue with your solution is to consider a very similar question:

{2, 3}
{2, 4, 6, 8, 10}
{1, 3, 5}
If one number is randomly selected from each set above, what is the probability that the product of the three selected numbers is even?

Since all five numbers in the second set are even, the product of the three selected numbers is guaranteed to be even, regardless of what happens with the selections from the other two sets.

### thank you ! that's clear now

thank you ! that's clear now

### In the last question why are

In the last question why are we not determining the probability of an odd number being selected in the first set, second set and the third set. ### We can use counting methods

We can use counting methods or probability rules to solve this question. In the video, we use counting methods, because we haven't yet covered probability rules in the module.

Using probability rules, we can say the following:

P(product is odd) = P(# from 1st set is odd AND # from 2nd set is odd AND # from 3rd set is odd)
= P(# from 1st set is odd) x P(# from 2nd set is odd) x P(# from 3rd set is odd)
= 1/2 x 1/5 x 1/4
= 1/40

So, P(product is even) = 1 - 1/40 = 39/40

Thanks.

### Sir how to solve this

Sir how to solve this question
A box contains 11 hats, out of which 6 are red hats and 5 are green hats. If two hats are to be selected at random without replacement, what is the probability that at least one green hat will be selected? ### The easiest solution is to

The easiest solution is to use the complement.

P(at least one green hat) = 1 - P(ZERO green hats)

Let's calculate P(ZERO green hats)

P(ZERO green hats) = P(1st hat is red AND 2nd hat is red)
= P(1st hat is red) x P(2nd hat is red)
= 6/11 x 5/10
= 30/110
= 3/11

So, P(at least one green hat) = 1 - 3/11
= 8/11

Cheers,
Brent

### hi Brent,

hi Brent,

How would you solve the product of 3 numbers question if you have 2 odd numbers in the 2nd group? You're referring to the question that starts at 2:30.
Let's change the numbers in the second set to include TWO odd numbers to get:
{2, 3}
{5, 9, 12, 18, 20}
{4, 6, 9, 10}

So, P(product is EVEN) = 1 - P(product is ODD)

P(product is ODD) = P(1st number is odd AND 2nd number is odd AND 3rd number is odd)
= P(1st number is odd) x P(2nd number is odd) x P(3rd number is odd)
= (1/2) x (2/5) x (1/4)
= 1/20

So, P(product is EVEN) = 1 - 1/20
= 19/20

Cheers,
Brent

### For the question below, I was

For the question below, I was thinking a group of siblings of 4 and another group 3 (I.e 4 +3) as opposed to 2+2+ 3

What is the hint to crack the 2+2+3. Based on solutions it seems it is obviously but I Could not see it. Is it answer choices?

https://www.beatthegmat.com/siblings-t280457.html Good idea, but "a group of siblings of 4 and another group 3" does meet the given condition that 4 people have exactly 1 sibling in the room.

If we have a family of 4 siblings, then each of those siblings will have 3 siblings in the room (not 1)

Cheers,
Brent

### Hi Brent,

Hi Brent,

Firstly apologies if I have been asking too many questions lately (as mentioned before, Counting/Probability are my weakest points).

Given this question: What is the probability of rolling three fair dice and having at least one of the three dice show an even number?

A. 35/36
B. 7/8
C. 1/8
D. 1/36
E. 1/216

Could you please explain why we need to find the complement in order to obtain the correction solution? I can't seem to understand the logic behind this given that each dice roll/probability set contains an equal number of odd and even numbers.

For instance, why can't we go straight to the conclusion that each roll has a 3/6 chance of giving us an even number? Why is that we must first calculate the probability of an odd roll (which is also 3/6) in order to find the complement then subtract from 1.

AKA if one die has 3 even and 3 odd numbers, this implies that we have a 1/2 probability to roll an even or odd number. If we are to roll this one die 3 times, we get:

P(EVEN) --> (1/2) x (1/2) x (1/2) = 1/8

To me, this logically answers the question being asked. But I can't seem to understand why it is wrong. Appreciate any clarification!

Thanks as always :) NEVER worry about asking a lot of questions. That's exactly what you're supposed to do when you're experiencing difficulties.

We want to determine P(at least one even)
If we don't use the complement here, we must consider all of the possible ways to get at least one even number.

So, we must consider:
- getting exactly one even number
- getting exactly two even numbers
- getting exactly three even numbers

In other words: P(get at least one even) = P(get 1 even number OR get 2 even numbers OR get 3 even numbers)
= P(get 1 even number) + P(get 2 even numbers) + P(get 3 even numbers)

As you can see, each of these calculations will take a lot of work.
However, if we use the complement (as I have done here https://gmatclub.com/forum/what-is-the-probability-of-rolling-three-fair...), we will have fewer calculations to make.

Important: When solving probability questions, it helps to rewrite the probability in terms that are clear to you (more here https://www.gmatprepnow.com/module/gmat-probability/video/754)

For example, above I rewrote the probability as follows:
P(get at least one even) = P(get 1 even number) + P(get 2 even numbers) + P(get 3 even numbers)

Then we should do the same for each of the above probabilities.

For example: P(get 3 even numbers) = P(1st roll is even AND 2nd roll is even AND 3rd roll is even)
= P(1st roll is even) x P(2nd roll is even) x P(3rd roll is even)
= (1/2) x (1/2) x (1/2)
= 1/8

I believe your solution, P(EVEN) = (1/2) x (1/2) x (1/2) = 1/8, is calculating the probability that all three rolls are even.

Cheers,
Brent

### Thank you so much! I'm

Thank you so much! I'm starting to piece this section together much better than counting. Also, thanks for always responding so diligently! ### Good stuff! Glad to help! ### Hi Brent,

Hi Brent,

Sorry my last qs was very stupid. I moved away from my desk and when I came back I re-read the qs. I guess doing so many qs at a time and not taking a break has made me tired and I overlook such simple details.

My apologies and sorry to have bothered you. ### No need to apologize. All

No need to apologize. All questions are welcome! :-)

Cheers,
Brent

### Hi Brent,

Hi Brent,

I solved the below question using complement but ain't getting the answer right. My method:

1 - 8C6/10C6 (8C6: Removed Alan and Becky from the group of 10 people). Where am I going wrong in this?

https://gmatclub.com/forum/alan-and-becky-are-among-the-ten-students-a-teacher-can-choose-from-to-285483.html To better understand the ramifications of this question, let's say that I'm the leader of a very elite club.
In order to be in this club, you must meet BOTH of the following requirements:
1) Your birth month must be April
2) You must be right-handed
If you don't meet both requirements, then you cannot be in the club.

Now, let's say Joe applies, but does not get to be in the club.
What can we conclude about Joe?
Can we conclude that he wasn't born in April AND he isn't right-handed?

No, there are 3 possible scenarios that disqualify Joe from being in the club:
- Joe was born in April, and Joe is left-handed
- Joe was not born in April, and Joe is right-handed
- Joe was not born in April, and Joe is left-handed
-------------------------------

The same logic applies to the original question.

If Alan and Becky are both selected to be on the team, then Alan and Becky are BOTH on the team.
What if Alan and Becky are NOT both on the team?

There are 3 possible scenarios:
1) Alan was selected, and Becky was NOT selected
2) Alan was NOT selected, and Becky was selected
3) Alan was NOT selected, and Becky was NOT selected

You have accounted for scenario #3 only.
If we factor in scenarios 1 and 2, you'll arrive at the correct answer.

Cheers,
Brent

### https://gmatclub.com/forum/in

https://gmatclub.com/forum/in-a-room-filled-with-7-people-4-people-have-exactly-1-sibling-in-the-87550.html

There are two groups of siblings- (1234)
and (567). Total number of combinations of 2 students is 7C2=21.
Why can't I take total number of ways of selecting one sibling from each group as 4*3= 12
The various combinations are-
15
16
17
25
26
27
35
36
37
45
46
47
So 12/21 That's a good strategy, but you're missing four possible outcomes.
If 1&2 are siblings, and 3&4 are siblings, then there are 4 additional outcomes in which the two selected people are NOT siblings:
1) 13
2) 14
3) 23
4) 24

So, there are 16 allowable outcomes (not 12)

Cheers,
Brent

### Ohh yes! I totally missed

Ohh yes! I totally missed that part. Thank you!

### https://gmatclub.com/forum/8

https://gmatclub.com/forum/8-cities-including-memphis-are-finalists-to-be-chosen-to-host-a-poli-187957.html

In this question, statement 1 will not be helpful at all imo.

because it saying that one out of the 8 are selected at a 7/8 probability. I think you may be reading statement 1 incorrectly.

Target question: What P(Memphis is NOT chosen)?

Statement 1: The probability that any one of the 8 cities does NOT win the competition is 7/8.
In other words, for EACH of the 8 cities, P(that city is NOT chosen) = 7/8
Since Memphis is one of those 8 cities, we can conclude that P(Memphis is NOT chosen) = 7/8
So statement 1 is sufficient.

Does that help?

### Yes, got it. Thank you so

Yes, got it. Thank you so much!