Question: Sum of 11

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.

There's a nice solution here: http://www.urch.com/forums/gmat-problem-solving/37062-probability-averag...

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.

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

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

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.

Interesting solution! Nice work!

Essentially, your approach is as follows:
P(sum is 11) = P(2 then 9 OR 4 then 7 OR 6 then 5 OR 7 then 4)
= P(2 then 9) + P(4 then 7) + P(6 then 5) + P(7 then 4)
= (1/5)(1/8) + (1/5)(1/8) + (1/5)(1/8) + (1/5)(1/8)
= 4(1/5)(1/8)

Your solution is perfectly valid.
Here's how it breaks down mathematically:
P(sum is 11) = P(1st # is 1 of the 4 #'s that can create a sum of 11 AND 2nd # is the matching # to make 11)
= P(1st # is 1 of the 4 #'s that can create a sum of 11) x P(2nd # is the matching # to make 11)
= 4/5 x 1/8 = 1/10

As you can see, there's no duplication here.