# Question: At Least 1 Even Number

## Comment on At Least 1 Even Number

### In this question, the total

In this question, the total no. of possible outcomes is not 10c2, as if two no are selected randomly, these can be the outcomes (23,24,25,26,32,34,35,36,42,43,45,46,52,53,54,56,62,63,64,65), i.e 20 outcomes.However, the final answer will be the same ### I don't say that the total

I don't say that the total number of outcomes is 10C2. I say it's 5C2.

We have two options here. We can say that order does not matter in both the numerator and denominator (as I did in my solution). Or we can say that order DOES matter in both the numerator and denominator (as you are suggesting). As you say, however, the answer is the same in both instances.

### why is the answer for both

why is the answer for both order does matter and does not matter the same ### In most cases, we'll arrive

In most cases, we'll arrive at the correct probability as long as we treat the numerator and denominator the same way.

### can we answer it like:

can we answer it like: Probability of selecting 3 and 5=1/5* 1/4=1/20.

Therefore the ans =1-1/20=18/20.

is it possible? ### Close! First, 1 - 1/20 does

Close! First, 1 - 1/20 does not equal 18/20. So, we have a problem there :-)
Your calculation of (1/5)(1/4) = 1/20 represents the probability of selecting the 3 FIRST and the 5 SECOND.
We must also consider the probability of selecting the 5 FIRST and the 3 SECOND. This too is equal to (1/5)(1/4) = 1/20
So, P(selecting 3 and 5) = 1/20 + 1/20 = 1/10
So, P(selecting at least one even) = 1 - 1/10 = 9/10

### Can we solve this problem

Can we solve this problem this way?
We have to select two numbers (Num1 and Num2). There is "and" sign it , It means multiplication. if we select two odd numbers (2/5*1/4)=1/10. Then P(even No:) = 1-1/10=0.9.
Need you Acknowledgment for this procedure. ### Yes, that's a very valid

Yes, that's a very valid solution. This practice question appears early in the probability module and is meant to reinforce some of the more basic probability strategies.

### Hi,

Hi,
In your previous video about "p( product of 3 numbers will be odd)", you found out the total no: of outcomes in the denominator by choosing 3 no: from each of the 3 sets, which is 2x5x4, which could be interpreted as 2C1x5C1x4C1.

But why can't I use the same interpretation when it comes to choosing 2 no:s from a SINGLE set to get the denominator as 5C1x4C1 instead of 5C2? although I realize in this case order of the no's are inadvertently taken into consideration, can you give me a sound reasoning as to why that logic/method is forbidden? why am I not allowed to write here 5C1x4C1 while in the former I was able to split into 2C1x5C1x4C1? Is it because in the former it was 3 distinct sets and not allowed for a single set? ### Your question is similar to

We have two options here. We can say that order does not matter in both the numerator and denominator (as I did in my solution). Or we can say that order DOES matter in both the numerator and denominator (as you are suggesting). As you say, however, the answer is the same in both instances.

If you want to say that order matters when calculating the denominator, the we must say that order matters when calculating the numerator.

You have already calculated the denominator (with the premise that order matter), and you got 5C1 x 4C1 = 5 x 4 = 20

Now the denominator. In how many ways can we select 2 of the numbers so that we don't get any even numbers?
There are only 2 odd (i.e., non-even numbers).
So, there are 2 ways (2C1 if you wish) to select the first odd number, and there is 1 way to select the second odd number). So, in TOTAL, the number of ways to select 2 odd numbers = 2 x 1 = 2

So, P(select zero even numbers) = 2/20 = 1/10

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

### Why can't you do 2/5C2? Why

Why can't you do 2/5C2? Why does the numerator have to be 2C2? ### Please see my response to

Please see my response to ananthu's question above.

### Hi Brent,by following your

Hi Brent,by following your above video explanation, may I know if I can use 3C2 (for choosing two even numbers out of the three, instead of 2C2; for choosing the odd numbers) and divide it by 5C2.Thanks for clarifying. ### Hi Runnerboy44,

Hi Runnerboy44,

Yes, we can select 2 even numbers in 3C2 ways. However, I'm not sure what you want to do with this information.

The question asks us to find P(at least one even number)
There are two possible cases that give us at least one even number:
CASE 1: one number is even and the other number is odd
CASE 2: both numbers are even

So, 3C2 = the number of ways to accomplish CASE 2, but you still need to address CASE 1.

Does that help?

Cheers,
Brent

### yes it does help Brent. When

yes it does help Brent. When I evaluate for case 1, I get 3C1 x 2C1 = 3 x 2 = 6. So when I add it to case 2 ie 3C2 = 3, I get a total of 9 (ie 6+3). And 9/5C2 = 9/10 = 0.9 Perfect!

### Is how I did this sound?

Is how I did this sound?

possible ways to choose one even number, 5X3X2= 30 possible choices. 3 even numbers= 3/30= 0.1. P(not odd)= 1-0.1= 0.9. ### It's hard to tell whether

It's hard to tell whether this is a valid approach, but my gut feeling is that you arrived at the correct answer by coincidence. But I could be wrong.

What do 5, 3 and 2 represent in your calculation "5X3X2= 30"?
I ask because, there are 20 ways in which we can select 2 numbers (there are 5 ways to select the first number and 4 ways to select the second number. 5 x 4 = 20. So, we can select 2 numbers in 20 ways)

Also, when you write "3 even numbers= 3/30= 0.1"
What does 0.1 represent?

Cheers,
Brent

### Hey Brent, so i got the

Hey Brent, so i got the denominator for this problem with the combination problem 5C2. However is the the reason why you select 3, 5 as a pair and not individually due to the wording of the question, since they are asking for a selection? Originally i picked 0.8 as the answer since 3 and 5 are individual odd numbers. Any help is appreciated. Thanks ### It sounds to me like you

It sounds to me like you treated the numerator and denominator differently.

For the denominator, you used 5C2, which means order does NOT matter when selecting the two numbers.
For the numerator, you treated the outcomes as though order DOES matter. That is, you treated the outcome of selecting 3 first and 5 second as different from the outcome of selecting 5 first and 3 second.

If we're saying that order does NOT matter, then there's only 1 way to select at 3 and a 5.
So, P(both ODD) = 1/10

Please note that you would have calculated the same probability if we had treated the numerator and denominator as though order DOES matter.
When calculating the total number of outcomes as though order matters, there are 5 ways to select the first number, and then 4 ways to select the second number.
So the total number of outcomes = 5x4 = 20

Since we're saying that order matters this time, there are two different ways to select both the 3 and 5.
One way is to select 3 first and 5 second
The way is to select 5 first, and 3 second.
So, P(both ODD) = 2/20 = 1/10

Does that help?

### thank you that helped me .

thank you that helped me .