GRE/GMAT probability question

[Chriszzz]

Hi guys, first time poster here. I ran into some confusion while using Sid Thatte's GRE/GMAT math book. There are two probability practice problems which Mr. Thatte seems to solve with different approaches. They are as follows:

Question 1 (dice-sum problem). If you roll two dice, what's the probability that the sum of the two numbers you get will be 6?

Question 2 (list-product problem). If you have two lists of numbers, list A being 1-3-5 and list B being 2-5-6, you select one number from each list, what is the probability that the product of the two numbers will be odd?

My confusion is in the way the book calculates the denominator (aka the total # of outcomes) of each probability.

In the dice-sum problem, they counted the total # of outcomes as the number of dice roll arrangements, not as the total # of unique sums.

In the list-product problem, they did the exact opposite and counted the total # of outcomes as the number of unique products that were possible, not the total # of arrangements of number choices.

i.e., in the dice-sum problem, the book counts 36 total possible outcomes, which is 6*6, or the number of total arrangements/permutations of the two dice rolls. They did not use the total # of sums possible, which would be 11.

In the list-product problem, the book counts 8 total possible outcomes from the 8 possible products between the numbers within the two lists (2,5,15,1,10,25,30). They did not use the total # of numerical arrangements, which would be 17 (3*3*2, minus one because picking 5, then another 5 is the same thing as its reverse)

My question is, is this an error in the book or are the different approaches because of the different nature of the two problems (dice roll vs pick-a-number)? And if these were official ETS questions, how would I be expected to answer?

Thanks

[edgirl]

Yeah, I think the issue here is that, for example, a roll of 1 6 isn't the same as a roll of 6 1 -- they add up to the same sum, of course, but visually they're a different arrangement, and what you're being asked is how many successful arrangements (sum of 6) there are out of all the possible arrangements (36).

In the second problem, 2x3 is the same as 3x2 -- the outcomes aren't distinct -- and this is different from the first problem because you're being asked about possible products, not about the arrangement of factors that gets you to that product. That is, you're being asked about how many possible totals there are, not the ways you can get to the total. (I hope that makes sense!)

So I don't think it's a mistake, and I hope that's helpful. I'm sure other folks will jump in if they disagree with me!

[pemdas]

Hi guys, first time poster here. I ran into some confusion while using Sid Thatte's GRE/GMAT math book. There are two probability practice problems which Mr. Thatte seems to solve with different approaches. They are as follows:

Question 1 (dice-sum problem). If you roll two dice, what's the probability that the sum of the two numbers you get will be 6?

1-5, 2-4, 3-3 and vice versa 5-1, 4-2, 3-3. Total 6 options out of 6*6=36 possibilities. So the chance is p=6/36 or 1/6

Question 2 (list-product problem). If you have two lists of numbers, list A being 1-3-5 and list B being 2-5-6, you select one number from each list, what is the probability that the product of the two numbers will be odd?

In the list A we have all odd numbers and the list A contains only one odd number (5). The product of odd numbers is always odd, so we have only three options (5*1, 5*3, 5*5) out of possible 3*3=9 possibilities and the chance is p=3/9 or 1/3

My confusion is in the way the book calculates the denominator (aka the total # of outcomes) of each probability.

In the dice-sum problem, they counted the total # of outcomes as the number of dice roll arrangements, not as the total # of unique sums.

In the list-product problem, they did the exact opposite and counted the total # of outcomes as the number of unique products that were possible, not the total # of arrangements of number choices.

i.e., in the dice-sum problem, the book counts 36 total possible outcomes, which is 6*6, or the number of total arrangements/permutations of the two dice rolls. They did not use the total # of sums possible, which would be 11.

We are not interested in the possible sums of two rolls, as such you are correct – they would give us 1-6,2-6,… 6-6 and some 11 ( I don’t know I didn’t count). We are interested in the number of rolls of two dice to result in one event – the sum of 6.

In the list-product problem, the book counts 8 total possible outcomes from the 8 possible products between the numbers within the two lists (2,5,15,1,10,25,30). They did not use the total # of numerical arrangements, which would be 17 (3*3*2, minus one because picking 5, then another 5 is the same thing as its reverse)

My question is, is this an error in the book or are the different approaches because of the different nature of the two problems (dice roll vs pick-a-number)? And if these were official ETS questions, how would I be expected to answer?

In your book, the author has mistakenly discounted the products of 1*5 and 5*1 and counted this as one outcome, hence 8 possible outcomes, i.e. 9-1=8. But this is wrong as even with the same products the two numbers below to different sets hence both have their assigned probabilities to be picked up from the lists A and B. Let’s solve this questions differently,

Suppose we must have odd product which is possible under one single condition à odd*odd=odd. The probability that list A has an odd number is 1 or 100% and the probability that list B has an odd number is 1/3. hence 1*1/3=1/3. Now the vice versa, the probability that list B has an odd number is 1/3 and the probability that list A has an odd number is 1, hence (1/3)*1=1/3. We cannot add up these two probabilities (1/3 and 1/3) because the two lists A and B represent distinct pools and our selection for the product of numbers doesn’t DEPEND on our selection from either the set A first and set B next or the set B first and the set A next. To prove this we need to look into number theory and the distribution rule for multiplication à a*b=b*a. So we don’t over-count and leave only 1/3.