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Confused_111 · Oct 22, 2007, 02:15 AM

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

colbtech · Oct 22, 2007, 02:22 AM

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

Confused_111 · Oct 22, 2007, 02:30 AM

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

Capuchin · Oct 22, 2007, 02:36 AM

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

Confused_111 · Oct 22, 2007, 02:53 AM

No odd numbers out of the three roles??

Capuchin · Oct 22, 2007, 02:56 AM

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

Confused_111 · Oct 22, 2007, 03:09 AM

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

terryg752 · Oct 22, 2007, 03:16 AM

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

galactus · Oct 22, 2007, 08:23 AM

Agree with Terry.

By the binomial probability.

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