# Question: At Least 1 Even Number

## Comment on At Least 1 Even Number

### if we take two numbers in

If we take two numbers in which at least one is not even, it can be 3,5 and 5,3
It should be 2/10 = 0.2, please clarify. Your answer is 0.9! ### In your solution, you're

In your solution, you're treating the numerator as though order MATTERS (i.e., 3 then 5 is different from 5 then 3), BUT you are treating the denominator as though order does NOT matter (the number of outcomes = 5C2).

If you want to take the approach in which order matters (as you have done with the numerator), then we need to calculate the denominator differently.

If order matters, then there are 5 different ways to select the 1st number, and there are 4 different ways to select the 2nd number. So, the TOTAL number of ways to select two numbers = 5 x 4 = 20

For the numerator, you have already stated that there are 2 different outcomes:
1) 3 then 5
2) 5 then 3

P(zero even numbers) = 2/20 = 1/10
So, P(at least 1 even) = 1 - 1/10 = 9/10 = 0.9

### I don't get is 2c2 is 1 and

I don't get is 2c2 is 1 and ,5c2 is 10 . How do we calculate these and in which manner ### The following videos will

The following videos will help.

Introduction to Combinations: https://www.gmatprepnow.com/module/gmat-counting/video/787

Calculating Combinations: https://www.gmatprepnow.com/module/gmat-counting/video/789

Cheers,
Brent

### Can you please explain why

Can you please explain why you removed the even numbers when selecting the odd two numbers? ### You're referring to my

You're referring to my comments at 2:40 in the video.

At that point, we're trying to determine the number of ways to select 2 numbers such that BOTH numbers are odd.

If we ignore the even numbers, we can see that there are only 2 odd numbers to choose from. So, there's only one way to choose 2 odd numbers.

Does that help?

Yes thanks

### Hi Brent- any reason you are

Hi Brent- any reason you are assuming order of the selected number doesn't matter, if these would have been alphabets then I agree ordering doesn't matter? Also why we do get same answer either way? ### We can treat this question

We can treat this 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. Let's do it...

P(at least 1 even number) = 1 - P(ZERO even numbers)
= 1 - P(BOTH odd numbers)

NUMERATOR - # of ways to select both odd numbers.
For the first selection, we have 2 options (select 3 or select 5). Once we've made the first selection, we have 1 option (select the last remaining odd number).
So, the number of way to select both odd numbers = (2)(1) = 2

DENOMINATOR - # of ways to select any two numbers.
For the first selection, we have 5 options. Once we've made the first selection, we have 4 options remaining
So, the number of way to select two numbers = (5)(4) = 20

So, P(BOTH odd numbers) = 2/20 = 1/10

This means P(at least 1 even number) = 1 - 1/10 = 9/10

Does that help?

### I did initially solve this as

I did initially solve this as though order matters, because I assumed choosing 3 then 5 and 5 then 3 were two different outcomes. Is this just a coincidence that this problem results in the same answer if you assume order matters and if you assume it doesn't?

Is this the case for other problems like this where either approach will result in the same answer? ### Your approach is perfectly

Your approach is perfectly fine.

The key is that, if you're going to say that order matters, then you must calculate the numerator as though order matters, and you must calculate the denominator as though order matters.
As long as the numerator and denominator are treated the same (i.e., order matters, or order does not matter), you'll reach the correct answer.

### Brent,

Brent,

I did in a different way, could you check if it is a valid approach?

P (at least one even number) = 1 - P (0 even numbers) (equation 1)

P (0 even numbers) = 1st set x 2nd set (equation 2)
1st set we have 2 odd numbers in a total of 5 numbers, then 2/5
2nd set we have 1 odd number in a total of 4 numbers, then 1/4

Solving the equation 2:
P (0 even numbers) = 2/5 x 1/4 = 2/20 = 1/10 = 0.1

Solving equation 1:
P (at least one even number) = 1 - 0.1 = 0.9

Thanks,
Pedro ### Hi Pedro,

Hi Pedro,

That's a perfect solution.
This question appears before the video lessons on "AND" probabilities, so I only cover the solution that involves counting techniques in the above video.

Cheers,
Brent

### Hi Brent,

Hi Brent,

Is it correct to say the below. There are three possible draws that could accomplish the goal of 1 even

Even Even = 3/5 * 2/4 = 6/20
Odd Even 2/5 * 3/4 = 6/20
Even Odd = 3/5 * 2/4 = 6/20

6/20 + 6/20 + 6/20 = 9/10 ### That's a perfectly valid

That's a perfectly valid solution. Nice work!

### I am a bit confused. Odd Even

I am a bit confused. Odd Even isnt that the same as Even Odd. Why does ORDER matter? How to know when to use the order and when not to use the order? thanks ### We can use either approach

We can use either approach (order matters or order doesn't matter).
What matters most is that we use the same counting approach for both the numerator and the denominator

That said, another thing that matter is that the outcomes must be equally likely.

ORDER MATTERS SOLUTION
P(at least 1 even) = 1 - P(NO evens)
= 1 - P(both ODD)

Let's use counting methods to calculate P(both odds)

Denominator: There are 5 ways to choose the 1st number, and there are 4 ways to choose the 2nd number.
So, the denominator = 5 x 4 = 20

Numerator: There are 2 ways to choose an ODD number 1st, and there is 1 way to choose and ODD number 2nd.
So, the numerator = 2 x 1 = 2

So, P(both ODD) = 2/20 = 1/10

P(at least 1 even) = 1 - 1/10 = 9/10
---------------------------------

ORDER DOESN'T MATTER SOLUTION
P(at least 1 even) = 1 - P(NO evens)
= 1 - P(both ODD)

Let's use counting methods to calculate P(both odds)

Denominator = 5C2 = 10 (aside: all 10 outcomes are equally likely)

Numerator = 2C2 = 1 (we must choose 2 ODD numbers from the 2 available ODD numbers.

So, P(both ODD) = 1/10

P(at least 1 even) = 1 - 1/10 = 9/10
---------------------------------

PROBABILITY RULES SOLUTION
P(at least 1 even) = 1 - P(NO evens)
= 1 - P(both ODD)

Let's use PROBABILITY RULES to calculate P(both odds)
P(both ODD) = P(1st number is odd AND 2nd number is odd)
= P(1st number is odd) x P(2nd number is odd)
= 2/5 x 1/4
= 1/10

P(at least 1 even) = 1 - 1/10 = 9/10

---------------------------------
If I were to generalize, I'd suggest two approaches:
1) If you have the option of using counting methods or probability rules to solve a probability question, the better/faster/more accurate approach is usually to use probability rules
2) If you are going to use counting methods to calculate the numerator and denominator, it's often better (if possible) to do so as though order matters.

Cheers,
Brent

### Brent, how are you?

Brent, how are you?

Is the folowing logic valid? (consider that I'm working with complement)

I can create the denominator in two stages: For the first number I have 5 possibilities and for the second 4, which makes a total of 20 possibilities.

As per the numerator, I can also create it in two stages. The first I have 2 possibilities and for the second I have just one, which makes a total of 2 possibilites.

So, when I divide 2/20 I end up with 0.1, so that I have a 0,9 chance to have at least one even number selected (complement).

Is that correct?

Thank you. ### I'm doing well, Pedro -

I'm doing well, Pedro - thanks!
Your solution is perfect; nicely done!

Cheers,
Brent

### Hi Brent, could we use the

Hi Brent, could we use the FCP to calculate this with the NON complement approach, ie the direct approach?

(Even and Odd + Even and Even) / 5C2

= (3 *2 + 3*2) / 10 .....first there are 3 evens to pick from, then there are 2 odds to pick from. Thats one outcome. The second outcome is first there are 3 evens to pick from and then there are ONLY 2 remaining events to pick from. What am I doing wrong here? ### You are treating the

You are treating the numerator as though the order of the selections matters, BUT you are treating the denominator as though the order of the selections does NOT matter.

Also, you are missing one type of outcome: odd on 1st selection and even on 2nd selection.

NUMERATOR
So, there are 3 ways to get at least one even
1) 1st is even AND 2nd is odd (3 x 2 = 6)
or
2) 1st is even AND 2nd is even (3 x 2 = 6)
or
3) 1st is odd AND 2nd is even (2 x 3 = 6)

Total = 6 + 6 + 6 = 18

DENOMINATOR
5 ways to select 1st number
4 ways to select 2nd number
Applying the FCP, we get: TOTAL = 5 x 4 = 20

So, P(at least one even) = 18/20 = 9/10 = 0.9

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

I used the probability rules to solve this question.

1- [P(odd number) + P(2nd odd number)]
=1 - [2/5 + 1/4 ( cos without replacement)]
= 1- 13/20
= 7/20

where am i going wrong here ? ### To use the probability rules

To use the probability rules here, we must recognize that your solution will involve an AND probability, and with AND probabilities, we must MULTIPLY the probabilities.

Here's my solution.

P(at least one even) = 1 - P(at ZERO evens)
= 1 - P(both selections are odd)

P(both selections are odd) = P(1st selection is odd AND 2nd selection is odd)
= P(1st selection is odd) x P(2nd selection is odd)
= 2/5 x 1/4
= 1/10
= 0.1

So, P(at least one even) = 1 - 0.1 = 0.9

Does that help?

Cheers,
Brent

### I solved it this way:

I solved it this way:

The probability that the event will happen the first time a number is picked is: 3/5

The probability that the event will happen the second time a number is picked is equal to the probability that an even is not picked first (2/5) times the probability that the even is picked second which is 3/4)

Then you add the two probabilities to get 18/20 which equals 0.90. ### That's a perfect solution.

That's a perfect solution.

Aside: this practice question appears before any video lessons on probability rules, so I only cover the solution that involves counting techniques in the above video.

Cheers,
Brent

### Hey Brent,

Hey Brent,

I found the answer without using combinations, and I want to just make sure my logic is valid.

Based on the wording of the question: select 2 random numbers from the set of {2,3,4,5,6} without replacement, whats the P that at least one number is even.

Given that there are two stages, and the numbers cannot be duplicates, for deciding on the denominator, I multiplied 5 x 4 = 20 possible two number selections (choose any one of 5 numbers for first number x choose anyone of 4 remaining numbers).

Then for the numerator, I identified the various ways ONLY ODD numbers can be selected: 3&5, and 5&3, so 2 total occurrences.

2/20 = 1/10 = 0.10

P(at least one even) = 1 - P (only odd) = 1 - 0.10
P(at least one even) = 0.90 = E

Just want to make sure my logic and reasoning behind my choices are correct.

Thanks again for you videos and teachings - they are very helpful. ### Your approach is perfect.

Your approach is perfect. Nice work!

### Hey Brent,

Hey Brent,
I see folks solving this using the Combination rule 7C2, but isn't possible to solve this using the MISSISSIPPI rule 7!/(2!)(2!)(3!) to get the denominator. But then im stuck on how to get the numerator.

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

We use the MISSISSIPPI rule to count the number of ways to arrange objects when some of those objects are identical.
So, for example, 7!/(2!)(2!)(3!) represents the number of ways to arrange 7 objects, where 3 of the objects are identical, 2 other objects are identical, and the remaining 2 objects are identical.

In the question you're referring to, we aren't arranging all 7 people; we're selecting 2 people from those 7 people.

Here's my solution that uses counting methods: https://gmatclub.com/forum/in-a-room-filled-with-7-people-4-people-have-...

Here's another solution that uses probability rules: https://gmatclub.com/forum/in-a-room-filled-with-7-people-4-people-have-...