Roll a die twice, what is the probability of getting at least 1 even number?

In my work, I have to show intermediate probabilities as well as the final answer. I put it below to give an idea of what's going on in my head!

1st roll: 1, 2, 3, 4, 5, 6

2nd roll: 1,1 2,1 3,1 etc. etc. etc.
1,2, 2,2 3,2
..... ..... ....
1,6 2,6 3,6

P(getting at least 1 even number)= P(getting 1 even number only) + P(getting 2 even numbers)

P(getting at least 1 even number)= 18/36 = 0.5
P(getting 2 even numbers)= 8/36= 0.22
0.5+0.22= 0.72 (or 26/32)

does that look A+??

I think the easier way to do this probability is to note that the complement of the event (getting at least one even number) is the event (getting NO even numbers). If we can find P(getting NO even numbers), then 1 minus that will give us P(getting at least one even number).

Of course, getting NO even numbers really just means both our numbers were odd.

The probability of getting an odd number on each roll is 1/2. Since the rolls are independent:

P(getting NO even numbers) = 1/2 * 1/2 = 1/4.

Thus, P(getting at least ONE even number) is 1 - 1/4 = 3/4.

__________________
Todd Werner
Director, Mathnasium West LA

You really must learn to simplify your life when possible.

Pr(at least one even number) = 1 - Pr(Only odd numbers)

Pr(only odd numbers) = (1/2)^2

Pr(at least one even number) = 1 - (1/2)^2

Your way:

Pr(even number then odd number) = (1/2)^2
Pr(odd number then even number) = (1/2)^2
Pr(even number then another even number) = (1/2)^2

Pr(at least one even number) = 3(1/2)^2

Unique answers do not care how you find them. The differenece is in the headaches.