If a is a number that is randomly selected from Set A, and b

Pick both +ve: 2/5 * 2/4 =1/5.
Pick both -ve: 2/5 * 2/4 = 1/5.
Required prob = 1/5 + 1/5 = 2/5.

Remember: "0" is neither +ve nor -ve.

HI All,

As far as probability questions are concerned, this one is fairly straight-forward. Even if you're not perfectly knowledgeable about all of the variations that probability questions can take and/or you might not be comfortable with all of the math 'steps' involved, you can still answer this question with a bit of 'brute force'...

A = {0, 1, -3, 6, -8}
B = {-1, 2, -4, 7}

Since Set A has 5 terms and Set B has 4 terms, there are (5)(4) different 'pairs' of numbers that can be selected. We're asked for the probability that the product of that pair is POSITIVE. We can simply list out the possible pairs that fit this restriction:

1 and 2
1 and 7
-3 and -1
-3 and -4
6 and 2
6 and 7
-8 and -1
-8 and -4

There are 8 pairs (out of 20 possible) that fit what we're looking for. 8/20 = 2/5

Final Answer:

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

Hello everyone!

I understood the problem but, does anyone know why at the end we need to sum instead of multiply the possibilities of +ve and -ve?

P = 4/20 + 4/20 = 8/20 = 2/5

Kind regards!

Hi jfranciscocuencag,

When a question asks for a TOTAL, then addition is always a possible 'approach.' Multiplication is SOMETIMES a possible approach (depending on what you're asked for and how the various pieces of information relate to one another).

A = {0, 1, -3, 6, -8}
B = {-1, 2, -4, 7}

Here, Set A has 5 elements and Set B has 4 elements. The TOTAL number of pairs (meaning 1 element from Set A and one element from Set B) can be determined either by listing out every possible pair (re: addition) or by multiplying 5 and 4 (since the nature of the question requires all possible pairs, multiplication is an option since it leads to that total result).

The total probability that we're asked to solve for here requires that we define 2 DIFFERENT groups (every pair of 2 POSITIVE terms and every pair of 2 NEGATIVE terms). Obviously, addition is an option - since we can total up those DIFFERENT options and get a grand TOTAL. Those two groups have NOTHING to do with one another though, so there's no logical reason to multiply those results (and doing so would NOT lead to the correct answer.

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

Thank you EMPOWERgmatRichC !

Now is clear for me.

The number a is selected randomly from Set A; the number b is selected randomly from Set B.
Therefore, total number of outcomes = \(5_C_1\) * \(4_C_1\) = 20. The denominator of the required probability will be 20.

This helps us eliminate answer options B and D since the number 20 can never be simplified to yield 3 or 9 since neither of these are factors of 20.

To find out the probability that the product of ab is positive, we can use the strategy,
Probability (ab > 0) = 1 – Probability (ab ≤ 0)

Probability (ab ≤ 0) = Probability (ab < 0) + Probability (ab = 0)

The product ab can only be made zero by making a = 0 and multiplying it with any of the four numbers in Set B.
Therefore, favourable outcomes for (ab = 0) = 4

The product ab can be made negative by making one of the numbers positive and the other number negative.

a can be – 3 or -8 and b can be 2 or 4; therefore, a total of 4 cases where the product ab is negative.

a can be 1 or 6 and b can be -1 or -4; therefore, a total of 4 cases where the product ab is negative.

Therefore, favourable outcomes for (ab < 0) = 8

Hence, Probability (ab ≤ 0) =\(\frac{ 8 }{ 20}\) + \(\frac{4 }{ 20}\) = \(\frac{12 }{ 20}\) = \(\frac{3 }{ 5}\) and so,

Probability (ab > 0) = 1 – \(\frac{3 }{ 5}\) = \(\frac{2 }{ 5}\).

The correct answer option is C.

I solved as below:

Total possibility - 5 * 4 = 20

Then I checked with each number in 1st set where we get ab = 0 or ab < 0 to eliminate possibilities from 20

For 0 - we get 4 such cases (0 multiplied by all 4 numbers from Set B yield a zero which is ab = 0)
For 1 - we get 2 such cases
For -3 - we get 2 such cases
For 6 - we get 2 such cases
For -8 - we get 2 such cases

In total, we have 4+2+2+2+2 = 12 cases where product ab = 0 OR ab < 0.

We are left with 8 cases out of 20 where we would get ab > 0

Hence, 8/20 = 2/5 and c is the answer