Hi there this question really stumped me...

A die is rolled three time. The probablity of getting at least one odd number is what? A tree diagram would be too long to do and the lattice might be useful, other than that I have no clue on how to do it.

Thanks

Evens? (1:1) Each roll offers the same odds, 3 odd number 3 even numbers.

I'm sorry, I am still a bit lost Iget what your trying to say. Do I cube root it or something.
Thanks

What's the probability of getting no odd numbers? Therefore, what's the probability of getting at least one odd number?

No odd numbers out of the three roles??

Yes, that should be very simple to work out, then you just work out the probability of NOT getting that outcome.

Oh I see, its so simple. Would it be any chance by Pr of 7/8 of getting at least one odd number

How not to get any odd number:

Each time : must be from 2, 4, 6

Each time: Number of ways to get even no. = 3
Number of ways to get any number = 6

Each time dice is rolled, probability of even number = 3/6 = 1/2

Probability getting Only even numbers 3 times is (1/2)^3 = 1/8

Probability : At least one odd number: = 1 - 1/8 = 7/8

Agree with Terry.

By the binomial probability.

\sum_{k=1}^{3}C(3,k)(1/2)^{k}(1/2)^{3-k}=C(3,1)(1/2)^{1}(1/2)^{2}+C(3,2)(1/2)^{2}(1/2)^{1}+C(3,3)(1/2)^{3}(1/2)^{0}=\fbox{7/8}