# 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

The answer given is by list down all the possible outcomes and determine the answer.

Why my method is wrong??

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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 $p (A \cup B) = pA + pB - pA \ pB = \frac{1}{2}+\frac{1}{3}-\frac{1}{6} = \frac{2}{3}$.

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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 $P(B)$.

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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 $1-\frac13=\frac23$.

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