Jack has two dice, one has six equally probable sides, labeled 1, 2, 3, 4, 5, 6, and the other has seven equally probable sides, labeled 1, 2, 3, 4, 5, 6, 7. If Jack rolls both dice what is the probability that both of the numbers will be odd?

A. 3/14
B. 2/7
C. 1/3
D. 1/2
E. 12/21

Kudos for a correct solution.
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Porbability = Total number of acceptable cases / Total number of all the cases

In this case, total number of all the cases = 6*7 = 42

'Acceptable cases' =
1,1
1,3
1,5
1,7
Similarly with 3 and 5 for a total of = 3*4 = 12

Thus, the required probability = 12/42 = 2/7. Thus B is the correct answer.

Hi All,

This question is a fairly standard probability question, so figuring out the individual probabilities and multiplying them together is a great approach for this prompt (and if you're not comfortable with that math yet, FireStorm's 'brute force' approach to list out all the possibilities is ALSO great).

The answer choices are 'spaced out' enough that you can still get the correct answer with a bit of logic, but you do have to understand fraction-to-decimals conversions and the basic math/logic behind rolling standard dice.

Since a standard 6-sided die has 3 odd 'sides' and 3 even 'sides', the probability of rolling an odd number on a single die is 1/2. By extension, the odds of rolling two odd numbers on two dice is (1/2)(1/2) = 1/4. Here though, the second die is a 7-sided die and a little MORE than half of the numbers are odd (4/7 to be precise, but it's enough to say a 'little more than half'), so rolling a 6-sided die and a 7-sided die gives us a probability of getting two odds that is a little MORE than 1/4....

There's only one answer that's a little more than 1/4...

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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I think the 800Score official solution is too complex.

I'd just take the individual probabilities of rolling an odd number, and multiply them.

1/2 * 4/7 = 4/14 = 2/7

Answer: B