Probability of at Least One 6 in 5 out of 10 Rolls of Two Fair Dice

[mateomy]

If two fair dice are rolled 10 times, what is the probability of at least one 6 (on either die) in exactly five of these 10 rolls?

So this problem is hard to wrap my head around. I'm probably wrong on many counts, here's what I'm doing:

Two fair dice are rolled 10 times but this question only cares about 5 of them. Because there's two dice I have,
<br /> \frac{1}{6} \frac{1}{6} = \frac{1}{36}<br />
for my probability of getting a 6.

Because they're asking 'at least one 6', I feel it's appropriate to take the complement.
<br /> 1 - \sum_{i = 0}^{1} \left(5 \choose i \right)\left(\frac{1}{36}\right)^{i} \left(1 - \frac{35}{36}\right)^{5 - i}<br />

However, when I do this my answer is off by about 0.03. So I know something isn't right. I'm not even sure if I'm attacking this correctly. Need clarification. Thanks.

 

[CompuChip]

It's a pretty complex problem because it requires you to put some simpler pieces together.
Let's start by finding the probability that rolling two dice gives at least one 6. The probability of both coming up as 6 is 1/36. What is the probability of at least one (i.e. one or both) being a 6?

 

[mateomy]

Wouldn't that just be 34/36?

 

[mateomy]

Oops, I think I meant 1/18.

 

[CompuChip]

Now you're just guessing :-)
Remember: you will either have no sixes, or at least one. The probability of one of these happening is 1 (i.e. 100%). What is the probability of no sixes?

 

[mateomy]

Hmmm...

Well if, say, I just have one dice my probability of getting a six is 1/6...so my probability of getting no sixes must be 5/6. With two dice its just a multiplication so I'm saying that my probability of getting no sixes is 25/36.

 

[CompuChip]

Very good. Now can you use my other hint from post #5 to find the probability that there is at least one six in a single two-dice roll?

(Actually what I told you there has a name - it's the complement rule. Does that sound familiar?)

 

[mateomy]

By this multiplication I've been doing:

If I have a probability of getting a six as 1/6 and the probability of not getting a six as 5/6 then I have a probability of getting at most one six of 5/36. For at least one six I think I need to do the compliment of that so
1 - 5/36 or 31/36?

 

[Ray Vickson]

By this multiplication I've been doing:

If I have a probability of getting a six as 1/6 and the probability of not getting a six as 5/6 then I have a probability of getting at most one six of 5/36. For at least one six I think I need to do the compliment of that so
1 - 5/36 or 31/36?

Now you are just guessing and not thinking; relax, take a deep breath and go back to square one. The event "no sixes" is the same as the event "die 1 gives a non-6 AND die 2 gives a non-six". What is the probability that die 1 gives a non-six? Same question for die 2. Now what is the overall probability of no sixes in one toss? Can you now find the probability of at least one six in one toss?

 

[mateomy]

So my answer of getting no sixes (25/36) is incorrect?

 

[Ray Vickson]

So my answer of getting no sixes (25/36) is incorrect?

It is correct; I was responding to your previous post, where you asked if the probability is 31/36. You gave several different answers, so it is not clear which one you meant!

 

[mateomy]

Ah, I see. Sorry for the ambiguity. I'm reading some websites and I'm starting to see some of my answers are wrong. I'm still confused on the problem though.