Q:

Probability with a coin and a die

This is for the GRE:
A fair coin is tossed once and a fair die with sides numbered 1, 2, 3, 4, 5, and 6 is rolled once. Let A be the event that the coin toss results in a head. Let B be the event that the roll of the die results in a number less than 5. What is the probability that at least one of the events A and B occurs?
How do you solve this to come up with $\frac{5}{6}$ (correct answer)? Probabilities are not easy. Any tip you can give is appreciated. Thanks.

A:

To find the probability that at least one occur, we can find the probability that NEITHER occur and then subtract that from $1$.
$A$ not happening = coin is tail
Probability of $A$ not happening is $\displaystyle \frac{1}{2}$
$B$ not happening = die roll is $5$ or $6$
Probability of $B$ not happening is $\displaystyle \frac{1}{3}$
Probability of $A$ and $B$ both not happening is $\displaystyle \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$
Probability of at least one of $A$ or $B$ happening is $\displaystyle 1 - \frac{1}{6} = \frac{5}{6}$.

A:

The definition of 'the probability of at least one of A or B' is:
$$P(A ~or~ B) = P(A) + P(B) - P(A ~and~ B)$$. 
And since $A$ and $B$ are independent, 
$$P(A ~and~ B) = P(A) \cdot P(B)$$
So, 
\begin{align*}
P(A ~or~ B) &= \frac{1}{2} + \frac{4}{6} - \frac{1}{2} \cdot \frac{4}{6} \\
            &= \frac{7}{6} - \frac{2}{6} \\
            &= \frac{5}{6} 
\end{align*}

A:

If you're finding the probability calculations hard, it might help to make a table. That is usually quite helpful for these kinds of questions. Everything is equally likely, so the probability for each outcome is $1/12$:
    Head   Tails
1   1/12   1/12
2   1/12   1/12
3   1/12   1/12
4   1/12   1/12
5   1/12   1/12
6   1/12   1/12

You want either heads or a number less than 5. That is everything that's not in the (quite ugly) bottom right box: 
    Head     Tails
1   1/12     1/12
2   1/12     1/12
3   1/12     1/12
4   1/12     1/12
          --------
5   1/12  |  1/12
6   1/12  |  1/12

If you sum them, you get $10/12=5/6$. Or, equivalently, if you sum the bottom right box and subtract that from 1 (since there are no other possible outcomes) you also get $5/6$.

