1.1.2 Adding probabilities

Consider now the following slightly more complicated problem. Assume again that you toss the coins aa and bb. What is the probability that both coins show the same outcome? There are two ways this could happen – either both coins are “heads” or both coins are “tails”. We already know from Eq. 1.14 that the probabilities of these two individual events are

p00=a0⁢b0,p11=a1⁢b1.\displaystyle p_{00}=a_{0}b_{0},\qquad p_{11}=a_{1}b_{1}.

Then the probability that one of these two events occurs is obtained by adding the probabilities:

p00+p11=a0⁢b0+a1⁢b1.p_{00}+p_{11}=a_{0}b_{0}+a_{1}b_{1}. (1.15)

You add probabilities when you want to group multiple outcomes of the same experiment together. Such combined events can usually be described using the word “or”. For example, “both coins are heads or both coins are tails”. Be careful: this only works when the two coins do not influence each other!

Exercise 1.3 (Adding probabilities).

Bob is also bored during the math class. He notices that Alice is staring at her watch so he takes a look at his watch too. Surprisingly, its second’s counter shows 44, which seems very unlikely to Bob. What is the probability that both digits of the second’s counter are the same if Bob takes a look at his watch at some random point within a minute?

Solution. There are six possible cases when both digits are the same (from 00 to 55). Each of these cases occurs with probability 160\frac{1}{60}. We can group them together in a single event whose probability is the sum of the probabilities of the six individual events:
160+160+160+160+160+160⏟6 terms=660=110.\underbrace{\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+% \frac{1}{60}}_{\text{$6$ terms}}=\frac{6}{60}=\frac{1}{10}.

Now that you know when to add and when to multiply probabilities, try to solve your first homework assignment!

Homework 1.1 (Opposite coins).

Alice tosses two coins called aa and bb, with probability distributions

a=(2/31/3),b=(3/41/4).\displaystyle a=\begin{pmatrix}2/3\\ 1/3\end{pmatrix},\qquad b=\begin{pmatrix}3/4\\ 1/4\end{pmatrix}.

What is the probability that the two coins produce opposite outcomes?

Hack.

Similar to Eq. 1.15, we need to compute p10+p01p_{10}+p_{01}. The probability is

p10+p01=a0⁢b1+a1⁢b0=23⋅14+13⋅34=512.\displaystyle p_{10}+p_{01}=a_{0}b_{1}+a_{1}b_{0}=\frac{2}{3}\cdot\frac{1}{4}+% \frac{1}{3}\cdot\frac{3}{4}=\frac{5}{12}.