Dice "or" probability question.

[DJMath]

I was thinking about the following problem: What are the odds you roll a die and it comes up with either an odd number or a prime number?

In my head:
Odd: 1,3,5
Prime: 2,3,5
The die could come up 1, 2, 3, or 5 and fit the criteria of being odd or prime, meaning 4/6

However, I was thinking about the following formula: P(A or B) = P(A) + P(B) - P(A and B) which would mean: 1/2 + 1/2 - (1/2 * 1/2) = 3/4. What am I missing? Thoughts? Thanks.

 

[R.M.]

P(A and B) = 1/3. A and B share 2 numbers out of the 6, so the probability that you roll a number that is in both sets is 2/6 = 1/3.

 

[DJMath]

P(A and B) = 1/3. A and B share 2 numbers out of the 6, so the probability that you roll a number that is in both sets is 2/6 = 1/3.

Thanks. I understand that that P(A and B) is 1/3 from the first way I went about solving it. However, using the formula (and maybe I'm using the wrong formula or missing something) P(A and B) = P(A)*P(B) which would be (1/2)*(1/2) = 1/4 which would mean P(A or B) = P(A) + P(B) - P(A and B) which would mean: 1/2 + 1/2 - (1/2 * 1/2) = 3/4. which is how I got the 2nd way of solving.

 

[pka]

Thanks. I understand that that P(A and B) is 1/3 from the first way I went about solving it. However, using the formula (and maybe I'm using the wrong formula or missing something) P(A and B) = P(A)*P(B) which would be (1/2)*(1/2) = 1/4 which would mean P(A or B) = P(A) + P(B) - P(A and B) which would mean: 1/2 + 1/2 - (1/2 * 1/2) = 3/4. which is how I got the 2nd way of solving.

You are using events [imath]A~\&~B[/imath] as if they were independent. Are they?
Does [imath]\mathcal{P}(A\cap B)=\mathcal{P}(A)\cdot\mathcal{P} (B)~?[/imath]