Question: Sum of 11

Comment on Sum of 11

Jamal.uddin's picture

Great

sir i have a doubt here we are selecting pairs in that one pair can be selected in alternate ways so total 8 sets ex (2+9),(9+2),,(4+7)(7+4),,(5+6)(6+5),,(4+7)(7+4) like this so 8/40=1/5. Whereas coming to the dice we do the same why not here can u explain please.
greenlight-admin's picture

Let's use your analogy with the dice to illustrate the problem. We have 2 dice, so let's make one die red and one die blue.

With this set-up, we can see that there are two ways to get a sum of 11:
case a: the red die is 5 and the blue die is 6
case b: the red die is 6 and the blue die is 5
IMPORTANT: The two cases above are DIFFERENT.

This set-up, however, does not apply to the question in the above video. Let's call the sets A and B.
A: {1, 2, 4, 6, 7}
B: {2, 3, 4, 5, 6, 7, 8, 9,}

Let's LIST the ways can we get a sum of 11:
case a: select 2 from set A, and select 9 from set B
case b: select 4 from set A, and select 7 from set B
case c: select 6 from set A, and select 5 from set B
case d: select 7 from set A, and select 4 from set B

These are the only ways to get a sum of 11.

Notice that case b and case d are different cases. In one case, we select the 4 from set A and in the other case, we select the 4 from set B.

In your question, you are suggesting that 2+9 and 9+2 are DIFFERENT.
However, there's only ONE way to get a sum with a 9 and a 2: We MUST select a 2 from set A and select a 9 from set B, since it's impossible to select a 9 from set A and select a 2 from set B.

Hi,

I wanted to ask why are the two cases different, meaning why can't we have a possibility of selecting 9 from set B and 2 from set A?
Is it because first we have to select a number from set A and then one from set B and keep the order?

Thanks

greenlight-admin's picture

We could have arranged our solution by selecting a number from set B before selecting a number from set A, but the answer would be the same.

The key is to recognize that approaches yield the same outcomes.

For example, notice that the outcome created by selecting 9 from set B and selecting 2 from set A is the same outcome by selecting 2 from set A and selecting 9 from set B.

Does that help?

Cheers,
Brent

Three positive integers are selected at random from 9 consecutive positive integers. What is the probability that their average is also an integer?

a) 1/28

b) 9/29

c) 9/28

d) 1/3

e) 5/14

I have missed the source but the official answer provided to the question is 5/14.

Hello Brent,

How is the pair {4,7} different from {7,4}?
Should we not consider as redundant?
greenlight-admin's picture

Selecting a 4 from set A and a 7 from set B is different from a 7 from set A and a 4 from set B.

Was very clear, thank you. I now understand that 4,7 is not the same thing as 7,4. This was confusing me.

But I have another question. If we have N1 & N2 in (N1, N2), would (N1=4, N2=4) be the same as (N2=4, N1=4)? I mean is each pick different, even if you pick the same number from the two?

Thanks.
greenlight-admin's picture

Yes, selecting 4 from set A and selecting 7 from set B is DIFFERENT FROM selecting 7 from set A and selecting 4 from set B.

However, selecting 4 from set A and selecting 4 from set B is THE SAME AS selecting 4 from set B and selecting 4 from set A.

Cheers,
Brent

Hi Brent, is there anyway we could have used the combination to get the denominator, since the order of combination does not matter
greenlight-admin's picture

We are choosing only ONE number from each set, the order definitely doesn't matter.

We can select one number from the first set in 5C1 ways (= 5 ways). Or we can just recognize there are 5 ways to select one number from the first set.

Likewise, we can select one number from the second set in 8C1 ways (= 8 ways). Or we can just recognize there are 8 ways to select one number from the second set.

So, the number of ways to select one number from EACH set = (5)(8) = 40

Does that help?

Cheers,
Brent

I found the question a bit confusing at first. You should have included the word "then" after "and," so that we know that a number is being picked from the first set FIRST. Anyway, I found the denominator of 40 without using combinations. I reasoned that picking 1 number from the first set was a 1/5 chance, and then 1 number from the next set would be a 1/8 chance. I multipled those fractions together to get 1/40. Then there were 4 pairs that add up to 11. 4 times 1/40 = 1/10
greenlight-admin's picture

It's important to note that the order in which we choose two numbers makes no difference to the solution. Likewise, choosing the two numbers simultaneously will also yield the same solution.

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