If two different numbers are selected at random from the numbers 1, 2

Question

If two different numbers are selected at random from the numbers 1, 2, 3, 4, and 5, what is the probability that their product will be odd?

A. 1/2
B. 1/12
C. 1/20
D. 3/5
E. 3/10

gmatophobia · Answer

Both the numbers must be odd for the product to be odd.

Probability of the first number to be odd = 3/5

Probability of the first number to be odd = 2/4

Net probability = 3/5 * 2/4 = 3/10

Option E

shrutijain12 · Answer

I take (3/5)*(2/4)*2!, then why is this wrong?

AllisonBell · Answer

Hi!

It looks like the issue is that you have the first two probabilities correct (3⁄5 probability of getting an odd number for the first selection; 2/4 for the second), but there is no need to multiply those by 2!.

I suspect that you multiplied by 2! in an attempt to account for the two different orders the numbers could be multiplied. For example, if I selected 3 and 1, I could multiply 3*1 or 1*3. But both of those possibilities are already accounted for by the two probabilities we multiplied. There’s a 3⁄5 probability I’ll pick a 1, 3, or 5, and a 2/4 probability I’ll pick one of the other two numbers. That encompasses the possibility of picking 1 first and 3 second, or 3 first and 1 second.

I hope that helps! Let me know if I’m misunderstanding your reasoning. For more practice problems, check out ManhattanPrep’s   Free GMAT Starter Kit  , which includes problems and explanations written by our teachers.

Happy studying!

Ally Bell 
 ManhattanPrep GMAT Instructor

RahulSax · Answer

Total Case - 10
Odd Product - odd x odd - 03
Probability - 3/10

vasu1104 · Answer

here both numbers must be odd.

for first num to be odd, probability = 3/5 since there are 3 odd num to choose from for first position.
now for second num be odd, probability = 2/4, since there are only 2 odds are available and also only 4 total num are available.

so 3/5 * 2/4 = 3/10

choice E

FreshStove · Answer

Odd should be multiplied with Odd to get "Odd" number.
And we know probability of such outcomes is (1,3), (1,5) and (3,5)
Favorable outcomes = 3*2 = 6 (multiplied by 2 as the sequence of these numbers can be swapped)
Total outcomes = 4*5 = 20
So Probability = 6/20 = 3/10.

If two different numbers are selected at random from the numbers 1, 2

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If two different numbers are selected at random from the numbers 1, 2

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