If one number is to be chosen at random from the integers 0 through 9

Chosing First 

Even || Odd
0.5   || 0.5

Chosing Second

Even -> Even 0.5*0.5 (Result Even)
Even - > Odd 0.5*0.5 (Result Even)

Odd - > Even 0.5*0.5 (Result Even)
Odd - > Odd 0.5*0.5 (Result Odd) 

So for even 0.25 + 0.25 + 0.25 = 0.75

 

Hello guys!

I don't know if this is a rule, but I got to D this way:

0 1 2 3 4 5 6 7 8 9 

0* any of 10 = 100% Even
1* any of 10 = 50% Even
2* any of 10 = 100% Even
3* any of 10 = 50% Even
4* any of 10 = 100% Even
5* any of 10 = 50% Even
6* any of 10 = 100% Even
7* any of 10 = 50% Even
8* any of 10 = 100% Even
9* any of 10 = 50% Even

Total possible: 1000%
All Possibilities to EVEN: 750% = 3/4 of possible

Am I right?

Thank you!

 ­

 

­Solution

Scenario 1

The probability of selecting two odds = 1/2 * 1/2 = 1/4

Scenario 2

The probability of selecting two evens= 1/2 * 1/2 = 1/4

Scenario 3

The probability of selecting one even and one odd = 1/2 * 1/2 = 1/4

If we add Scenario 1 + 2 + 3 we get  1/4 + 1/4 + 1/4 = 3/4

Hence, option D

­

­How can a product of 2 Odd number be even?

­The probability of selecting any even no. from the range 0-9 is 5/10 which is 1/2 in either case. Let's consider that the even nos. from the list is 0,2,4,6,8, which is five favorable outcomes divided by the total outcomes which is ten. It's the probability of selecting the even nos. from the range 0-9 not the product of two even nos. I hope that I can make my point clear.

I believe I found a faster approach for this question.

The key for me was realizing that the product of two numbers from the sets given is always even UNLESS both of the numbers are odd.

Therefore, the probability that the product of two numbers from those sets is even will be 1- (Probability of selecting an odd number*Probability of selecting an odd number).

Since half of the numbers are odd, the probability of selecting an odd number is 5/10

So we have:

Probability that the product of two numbers from those sets is even= 1-[(5/10)*(5/10)]
Probability that the product of two numbers from those sets is even=1-[(1/2)*(1/2)]
Probability that the product of two numbers from those sets is even=3/4