# Probability with a coin and a die [closed]

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.

## closed as off-topic by Did, Najib Idrissi, mau, naslundx, user63181 Apr 29 '14 at 10:33

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## 3 Answers

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 $\displaystyle 1 - \frac{1}{6} = \frac{5}{6}$.

• Thanks, @2012ssohn. How did you know that you should subtract the probability of the events not occurring from 1? Is this a rule in probabilities? – Alex Apr 29 '14 at 2:22
• Yes. The probability of event $A$ is $1$ minus the probability of event $A$ not happening. More simply put, $P(A) = 1 - P(\sim A)$. – 2012ssohn Apr 29 '14 at 2:23
• Thanks. Any websites that can help with learning probabilities? – Alex Apr 29 '14 at 2:27
• Probability is actually one of my weaker areas, so you're probably asking the wrong person. :( I'm sure there are plenty of other people here who knows better than me though! – 2012ssohn Apr 29 '14 at 2:31
• @Alex I learnt probability from AOPS and KA. – Shaurya Gupta Apr 29 '14 at 5:43

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*}

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 $5/6$.