Question

You roll two dice. What is the probability that the sum of the dice is odd and one die shows a 5? A 6x6 table of dice outcomes will help you to answer this question.

A. 2/9
B. 1/3
C. 1/6
D. 23/36

I don't understand how to do this problem.

Answer 1

Tricky... What's the probability that at least one of the dice shows a 5?

Answer 2

Wait, I think I got it... 4/18?
4 5's out of 18 odds, which reduces to 2/9...

Answer 3

How do you get 4 5's out of 18 odds?

Answer 4

Wait, scratch what I said before. I don't understand how to do this.
It says a 6x6 table would help, but I keep thinking that I need to multiply something because of the "and."

Answer 5

Okay, well, let's consider two cases... but first, let's name the dice... the first and the second die.... sound good? :)

Answer 6

Yep!

Answer 7

Okay, so there are two cases... the first case is that 5 shows up on the first die and the second die shows an EVEN number.
(The non-5 die should have an even number so that the sum is odd)

Catch me so far?

Answer 8

Yeah, got it.

Answer 9

And the second case, predictably, is that the first die shows an even number and the second die has the 5.

Everything all right up to now? :)

Answer 10

Yeah.

Answer 11

Okay, so for the first case, what's the probability (for the first die alone) that it shows a 5?

Answer 12

5 being the sum?

Answer 13

No, just one die, what are the chances that it shows 5?
Just considering the first die alone for now...

Answer 14

1/6

Answer 15

That's right :)
And what are the chances for the second die (also alone) to show an EVEN number?

Answer 16

1/2

Answer 17

Brilliant...
So now, what are the chances that the two dice show 5 (for the first) and an even number (for the second).

Remember that the dice are independent of each other, so you just *ehem* multiply their probabilities ;)

Answer 18

(1/6)*(1/2)=(1/12)?

Answer 19

that's right :)
And that's just for the first case.

For the second case, it's just in reverse, but it should get the same probability: 1/12

Because the probability for the first die showing an even number is 1/2
and the probability for the second die showing 5 is 1/6
\[\large \frac12 \times \frac16 = \frac1{12}\]

Right? :)

Answer 20

I thought so as well, but 1/12 isn't one of the answer choices, which is why I'm confused. :/

Answer 21

Of course it isn't :)
1/12 is just the probability for either case 1 or case 2.

Btw, case 1 and case 2 are the ONLY ways you can make that scenario happen, right?

(IE, if you want a 5 and the sum to be an odd number, you either have 5 in the first die or in the second)

Answer 22

I see, thank you for your help! I'm in a bit of a rush, and I have to move onto other problems now, but I definitely will get back to this afterwards.

And I just have to say this: I'm pretty competent with algebra and all, it's just the dice and card problems that make me feel like I'm reading a foreign language haha.

Answer 23

It's just one more step :)

Answer 24

You just add the probabilities for both case 1 and case 2 (since they cannot both happen at the same time).
\[\Large \frac1{12} + \frac1{12}=\frac2{12}=\color{blue}{\frac16}\]

ta-daa :D

Answer 25

Oh, I see! It seems really simple NOW that you've thoroughly explained everything haha. Thank you very much for your help! :)

Answer 26

The actual explanation is far more detailed and tricky than that, but I hope you got the general idea...

By the way, this isn't the ONLY way to do this, there are other ways, but they ultimately lead to the same answer :)

Another way is to manually count all the possible rolls with a 5 and adding up to an odd number... :)

That's it, welcome to Openstudy :)

Answer 27

I think that was more than enough for now. Thanks again!! :)