[singh2010]

A three-faced die T has the numbers 1 to 3 on its faces. This die and a normal die C are rolled together.

A) What is the probability of scoring T = 1 and C = 3
B) What is the probability of getting an odd number with T and 2 with C?
C) What is the probability of the numbers on T and C either being both even or being both odd?

[Unknown008]

Come on, what did you do?

a) P(T = 1 and C = 3) = P(T = 1) x P(C = 3)

b) P(T = odd and C = 2) = P(T = odd) x P(C = 1)
= P(T = 1, 3) x P(C = 1)
= [P(T = 1) + P(T = 3)] x P(C = 1)

c) P(Both even or both odd) = P(Both even) + P(Both odd)
= P(T = even and C = even) + P(T = odd and C = odd)
= [P(T = even) x P(C = even)] + [P(T = odd) x P(C = odd)]

Did you notice that where there was 'and' I put 'x' and where there was 'or' I put '+'?

[singh2010]

Come on, what did you do?

a) P(T = 1 and C = 3) = P(T = 1) x P(C = 3)

b) P(T = odd and C = 2) = P(T = odd) x P(C = 1)
= P(T = 1, 3) x P(C = 1)
= [P(T = 1) + P(T = 3)] x P(C = 1)

c) P(Both even or both odd) = P(Both even) + P(Both odd)
= P(T = even and C = even) + P(T = odd and C = odd)
= [P(T = even) x P(C = even)] + [P(T = odd) x P(C = odd)]

Did you notice that where there was 'and' I put 'x' and where there was 'or' I put '+'?

So A would be 1/3 x 1/6 = 1/18?

B) (1/3 + 1/3) x 1/6 = 2/18 = 1/9

C) even - 1/3 x 3/6 = 3/18 = 1/6
odd - 2/3 x 3/6 = 6/18 = 1/3
together 1/6 + 1/3 = 3/6 = 1/2?

I know I have been asking a lot of questions recently, this is because my exam is soon and I am panicking lol. Thanks for the input you have given me over the last few days.

[Unknown008]

That's good :)!

You see, it's not that difficult! :)

Do not hesitate to ask, but show what you at least tried. You can also post what are the answers you got, we'll check whether they are correct or not. :)