how to solve "or" probability question Question: A coin is tossed and a dice is rolled simultaneously. find the probability that:
A head or a number greater than 4 are obtained
I done this: 1/2 + 2/6 =5/6
but the answer is 2/3.
The answer given is by list down all the possible outcomes and determine the answer.
Why my method is wrong??
 A: You have counted the probability that both happen twice.
You have two events $A$ is the event that a head comes up, and $B$ is the event that a 5 or 6 comes up. The events are independent presumably, so the inclusion-exclusion formula gives:
$p(A \cup B) = pA + pB - p(A\cap B)$. In your case, since the events are independent, you have $p(A\cap B) = pA \ pB$, so $p (A \cup B) = pA + pB - pA \ pB = \frac{1}{2}+\frac{1}{3}-\frac{1}{6} = \frac{2}{3}$.
A: Your method assumes both events can never happen together, i.e. you are assuming $P(A or B) = P(A) + P(B)$. However, the correct argument would have been $P(A or B) = P(A) + P(B) - P(A and B)$. You may think why we are subtracting $P(A and B)$. This is because we have already counted this twice : once in $P(A)$ and once in $P(B)$.
A: Let me do this a different way.  The probability of getting tails is $\frac12$.  The probability of rolling 4 or less is $\frac23$.  Both of these are independent of one another so the probability of getting tails and rolling 4 or less is $\frac12\times\frac23=\frac13$.  If you didn't get tails and a number less than 4 at the same time, then you either obtained a head or a number greater than 4.  So the odds of obtaining a head or a number greater than 4 is $1-\frac13=\frac23$.
