Three dice probability problem

[SEstudent22]

Hello, this is my first time posting a question on this forum. I have read the rules, but if I have broken any I apologize.

Here is the problem: "Three dice are thrown. What is the probability that on at least two of them, an even number appears?".

I realize that when dealing with "at least" problems, almost always the best solution is to work with the complement. But in this case it doesn't really matter, right? There are two ways I can get the result, find the probability that on exactly two dice there is an even number and on the last die there is an odd number, then add the probability that on all three dice, and even number appears. OR I can subtract from one, the sum of the probability that on all three dice, no even number appears and on exactly one die an even number appears and the other two have an odd number.
I can find the probability of all three dice having an even number and all of them having an odd number (the complement). But I get stuck while trying to figure out the "exactly two" or "exactly one" probability. I think I'm getting confused because the probability to get an even or an odd number (on all three dice) is the same.

The probability to get an even number on all three is 1/8. And the probability to get an odd number on all three is 7/8.

 

[pka]

[imath]6^3=216[/imath] is the number of triples made of [imath]\{1,2,3,4,5,6\}[/imath].
Now [imath]3^3=27[/imath] is the number of ways to get three odd numbers (all odds).
So what is the probability of all odds?

What is the probability of two odds and one even? Remember it can be [imath]EOO,~OEO\text{ or }OOE[/imath].

 

[Dr.Peterson]

I can find the probability of all three dice having an even number and all of them having an odd number (the complement). But I get stuck while trying to figure out the "exactly two" or "exactly one" probability. I think I'm getting confused because the probability to get an even or an odd number (on all three dice) is the same.

The probability to get an even number on all three is 1/8. And the probability to get an odd number on all three is 7/8.

Since the probabilities of even and odd are both 1/2, you could just make a list of the 8 equally likely outcomes: EEE, ... . In how many of these are there at least 2 evens?

 

[sna_gz]

this question seems to be tricky. especially if your mind has been working on the developed question, you might think this is one of them.
One important thing and also a good point is that order doesn't matter.
basically, we should just sum the number of ways which we have two even( 2 * 3) dice + everything which does not matter( 6 ) = 12
and the total possible probability is 18 ( 3 * 6).
so the answer would be 12/18 = 4/6.

 

[SEstudent22]

Since the probabilities of even and odd are both 1/2, you could just make a list of the 8 equally likely outcomes: EEE, ... . In how many of these are there at least 2 evens?

Thank you Dr.Peterson and pka. I realized the answer when I laid out all of the likely outcomes. 4 out of the 8 possible outcomes satisfy the requirement so that's why the answer is 4/8 or 1/2 (for at least two dice having even number). The way I found out that all of the dice are even is 27/216 (as pka said), with that logic I tried to get 4/8. I got lost trying to find an answer the hard way and I completely ignored the fact that I could lay out all of the possibilities and see the answer clearly...

 

[Dr.Peterson]

Thank you Dr.Peterson and pka. I realized the answer when I laid out all of the likely outcomes. 4 out of the 8 possible outcomes satisfy the requirement so that's why the answer is 4/8 or 1/2 (for at least two dice having even number). The way I found out that all of the dice are even is 27/216 (as pka said), with that logic I tried to get 4/8. I got lost trying to find an answer the hard way and I completely ignored the fact that I could lay out all of the possibilities and see the answer clearly...

My way is really a "trick" that works easily only because the events E and O turn out to be equally likely. So we do need to be able to handle the long way.

Here's help with that: There are 3*3*3 = 27 ways to get three evens (EEE); there are also 3*3*3 ways to get EEO, and likewise for EOE and for OEE. Just add them up.

this question seems to be tricky. especially if your mind has been working on the developed question, you might think this is one of them.
One important thing and also a good point is that order doesn't matter.
basically, we should just sum the number of ways which we have two even( 2 * 3) dice + everything which does not matter( 6 ) = 12
and the total possible probability is 18 ( 3 * 6).
so the answer would be 12/18 = 4/6.

I don't understand your thinking, but since your answer is wrong, the reasoning must be wrong as well.

One issue is that order does matter; in order to have equally likely outcomes, we must take into account which die has what value. There are 36, not 18, possible outcomes from two dice.