Problem Solving

rakeshd347 wrote:Mary and Joe are to throw three dice each. The score is the sum of points on all three dice. If Mary scores 10 in her attempt what is the probability that Joe will outscore Mary in his?

24/64
32/64
36/64
40/64
42/64

Please explain the answer.

First consider rolling 1 die.
If you were to roll a die millions of times, what would be the average value rolled?
Well, since each outcome (1,2,3,4,5 and 6) are all equally likely, the average of the outcomes will be 3.5 (since (1+2+3+4+5+6)/6 = 3.5)
Of course, it's impossible to roll 3.5, but notice that the 3.5 divides the outcomes into two parts. We have the numbers less than 3.5 (that is 1,2,3) and the numbers greater than 3.5 (that is 4,5,6).

Also notice that, in one roll, P(rolling less than 3.5) = 1/2, and P(rolling more than 3.5) = 1/2

Now consider rolling 3 dice.
If the average expected outcome is 3.5 when one die is rolled, the average expected sum will be 10.5 when three dice are rolled (since 3.5 + 3.5 + 3.5 = 10.5)

IMPORTANT: If 10.5 is the average expected sums, then half of all sum will be less than 10.5 and half will be greater than 10.5. In other words, P(sum is less than 10.5) = 1/2 and P(sum is greater than 10.5) = 1/2

The question asks us to find P(sum is greater than 10). This is the same as P(sum is greater than 10.5), which means this probability = [spoiler]1/2 = B[/spoiler]

Cheers,
Brent

parveen110 wrote:Hi Brent,

I'm sure there is a reason as to why the approach you suggested here is different from somewhat simpler approach posted by uva@90. I'm unable to figure out why. Please clarify.
Thanks!

Hi parveen110,

The great thing about probability questions it that they can often be solved using more than 1 approach.
Uva's approach is probably faster, but if a person doesn't spot that particular approach, mine also works.

Cheers,
Brent

Hi All,

Uva's approach was a bit "lucky" - it's based on the assumption that each outcome is equally likely (and that assumption is wrong); in other situations the logic would not hold. Here's why:

Each total that is possible from rolling 3 dice will not necessarily occur the same amount of ways as the others. Here's an example:

Ways to total 3:
1+1+1 = 1 way

Ways to total 4:
1+1+2
1+2+1
2+1+1 = 3 ways

Etc.

The reason Uva ended up with the correct answer is because the number of outcomes "balances out"

Ways to roll a 3 = Ways to roll an 18
Ways to roll a 4 = Ways to roll a 17
Ways to roll a 5 = Ways to roll a 16
Ways to roll a 6 = Ways to roll a 15
Ways to roll a 7 = Ways to roll a 14
Ways to roll an 8 = Ways to roll a 13
Ways to roll a 9 = Ways to roll a 12
Ways to roll a 10 = Ways to roll an 11

This means that the probability of rolling 3-10, inclusive = probability of rolling 11-18, inclusive

Since the question was asking for the probability of rolling > 10, that means 50% of the possibilities will "fit." That is the ONLY situation in which the number of totals is proportionate to the probability of outcomes.

If the question had asked for the probability of rolling above 11, then the answer would NOT have been 7/16.

GMAT assassins aren't born, they're made,
Rich

Hi parveen110,

The "math" behind solving for the probability that the total would be greater than 11 would be layered but not impossible (you'd just have to list out the possibilities); as such, the GMAT would be unlikely to ask you to answer that question without giving you some built-in alternative approaches. For example, the answer choices could be such that it would be easy to eliminate most (if not all) of the wrong answers for being too big, too small or simply "impossible."

You should always keep in mind that GMAT questions are based on patterns. Finding the pattern behind all of those Quant and Verbal questions oftentimes allows a Test Taker to avoid some of the math/vocabulary and find a quicker approach to the correct answer.

GMAT assassins aren't born, they're made,
Rich