# Mutually exclusive events

Working my way through the following problem:

### Problem

Suppose that $$E$$ and $$F$$ are mutually exclusive events of an experiment. Show that if independent trials of this experiment are performed, then $$E$$ will occur before $$F$$ with probability $$\frac{ P( E)}{P( E) + P( F)}.$$

I have the following come up with the following solution:

### Solution

Since $$P( E^c) = P( F)$$ Therefore $$\frac{ P( E)}{ P( E) + P( F)} = \frac{ P( E)}{ 1 - P( F) + P( F)} = \frac{ P( E)}{ 1} = P( E)$$

But I am unsure if I am able to assume $$P( E^c) = P( F)$$ as a given?

As well, I am particularly confused by the answer in the solution manual which makes it's argument as follows:

### Solution Manual

If $$E$$ and $$F$$ are mutually exclusive events in an experiment, then $$P( E \cup F) = P( E) + P( F)$$. We desire to compute the probability that $$E$$ occurs before $$F$$ , which we will denote by $$p$$. To compute $$p$$ we condition on the three mutually exclusive events $$E$$, $$F$$ , or $$(E \cup F )^c$$. This last event are all the outcomes not in $$E$$ or $$F$$. Letting the event $$A$$ be the event that $$E$$ occurs before $$F$$, we have that

$$p = P( A|E) P( E) + P( A|F) P(F ) + P( A|(E \cup F )^c) P( (E \cup F )^c)$$

$$P( A|E) = 1$$

$$P( A|F) = 0$$

$$P( A|(E \cup F)^c) = p$$

since if neither $$E$$ or $$F$$ happen the next experiment will have $$E$$ before $$F$$ (and thus event $$A$$ with probability $$p$$). Thus we have that

$$p = P( E) + p P( (E \cup F)^c)$$

$$= P( E) + p (1 − P( E \cup F))$$

$$= P( E) + p (1 − P( E) − P( F))$$

Solving for $$p$$ gives

$$p = \frac{ P( E)}{ P( E) + P( F)}$$

as we were to show.

Specifically his statement

since if neither $$E$$ or $$F$$ happen the next experiment will have $$E$$ before $$F$$ (and thus event $$A$$ with probability $$p$$)

• They mean: If neither $E$ or $F$ happens on the first trial, then the game starts over.
– user940
Jun 22, 2012 at 2:34
• So we are able to treat the experiment as if only mutually exclusive events $E$ and $F$ exist and my solutions is valid correct? Jun 22, 2012 at 2:43
• No, that is a separate issue. In fact, there is no need to assume that $E$ and $F$ are exhaustive.
– user940
Jun 22, 2012 at 2:47

To determine the probability that $E$ occurs before $F$, we can ignore all the (independent) trials on which neither $E$ nor $F$ occurred, that is, $(E\cup F)^c$ occurred, since we are going to repeat the experiment until one of $E$ and $F$ does occur. So, look at the trial of the experiment on which one of $E$ and $F$ has occurred for the very first time. We are given that on this trial, the event $E \cup F$ has occurred. But, we don't yet know which of the two has occurred. So, given the knowledge that $E \cup F$ has occurred, what is the conditional probability that it was $E$ that occurred (and so $E$ occurred before $F$ since this is the first time we have seen either $E$ or $F$)?

$$P(E \mid (E \cup F)) = \frac{P(E(E \cup F))}{P(E \cup F)} = \frac{P(E \cup EF)}{P(E) + P(F) - P(EF)} = \frac{P(E)}{P(E)+P(F)}$$ since $P(EF) = P(\emptyset) = 0$.

Alternatively, let $G = (E\cup F)^c = E^c \cap F^c$ be the event that neither $E$ nor $F$ occurs on a trial of the experiment. Note that $P(G) = 1 - P(E) - P(F)$. Then, the event $E$ occurs before $F$ if and only if one of the following compound events occurs:

$$E, (G, E), (G, G, E), \ldots, (\underbrace{G, G, \ldots, G,}_{n-1} E), \ldots$$

where $(\underbrace{G, G, \ldots, G,}_{n-1} E)$ means $n-1$ trials on which $G$ occurred and then $E$ occurred on the $n$-th trial. The desired probability is thus

$$P(E ~\text{before}~ F) = P(E) + P(G)P(E) + [P(G)]^2P(E) + \cdots = \frac{P(E)}{1 - P(G)} = \frac{P(E)}{P(E)+P(F)}.$$

Your solution is incorrect. If $$E$$ and $$F$$ are mutually exclusive, it means that $$E \cap F = \emptyset$$, therefore $$F \subseteq E^c$$; and therefore, $$P(F) \color{red}{\le} P(E^c)$$.

You are not interpreting independent trials of the experiment correctly. It might be helpful to consider an example. Suppose you are rolling a biased 6-faced die. Let $$E$$ denote the event that 1 or 2 turn up and $$F$$ denote the event that 3 or 4 turn up. Continue rolling the die until either $$E$$ or $$F$$ occur. What is the probability that $$E$$ occurs before $$F$$, that is what is the probability that you get 1 or 2 before you get 3 or 4 (in the repeated rolls of the die).

The statement

since if neither $$E$$ or $$F$$ happen the next experiment will have $$E$$ before $$F$$ (and thus event $$A$$ with probability $$p$$)

means that if neither $$E$$ or $$F$$ happen, that is if 5 or 6 is rolled, we roll the die again. Since the rolls are independent, the probability of getting $$E$$ before $$F$$ in the future experiments is $$p$$.

Rant: This problem and its solution shows why students find probability confusing. The problem is stated very informally. In my opinion, a formal statement of the problem will remove some of the confuson.

Consider an experiment $$\mathcal E_1$$ with probability measure $$P_1$$. Let $$E$$ and $$F$$ be two events in $$\mathcal E_1$$. $$P_1(E)$$ denotes the probability that $$E$$ occurs in experiment $$\mathcal E_1$$. Similarly interpretation holds for $$P_1(F)$$.

Now consider another experiment $$\mathcal E_2$$, which represents infinite independent repetitions of the experiment $$\mathcal E_1$$. Let $$P_2$$ be the probability measure for events in $$\mathcal E_2$$.

Now consider an outcome $$\omega$$ of $$\mathcal E_2$$ that is a series of outcomes of $$\mathcal E_1$$. Let $$\tau_E$$ denote the first time $$E$$ occurs in $$\omega$$ (with $$\tau_E = \infty$$ if $$E$$ does not occur). Similarly, let $$\tau_F$$ denotes the first time $$F$$ occurs in $$\omega$$. Let $$A$$ denote the event (in $$\mathcal E_2$$) that $$\tau_E < \tau_F$$. The question is asking you to show that

$$\displaystyle P_{\color{red}2}(A) = \frac{ P_1(E) }{ P_1(E) + P_1(F) }$$

Thus, the question is asking you to compare two different experiments. When you write $$E^c \equiv F$$, you were thinking in terms of experiment $$\mathcal E_2$$; but $$E$$ and $$F$$ are not events in $$\mathcal E_2$$; they are events in $$\mathcal E_1$$.

• Since as you state in the context of your example > if neither $E$ or $F$ happen, that is if 5 or 6 is rolled, we roll the die again. Approaching the problem as if $E^c \equiv F$ is therefore valid then, no? Jun 22, 2012 at 3:02
• $E^c = \{3,4,5,6\} \not\equiv \{3,4\} = F$ Jun 22, 2012 at 3:12
• Yes but should ${5,6}$ occur we roll again, for the purposes of calculating the desired probability of this problem we disregard all events that do not exist in $E \cup F$ as they have no effect on the computation, therefore you are able to approach the problem as if $E^c \equiv F$, no? Jun 22, 2012 at 3:20
• Does my updated answer clarify this point? Jun 22, 2012 at 3:25
• The event that $E$ does not occur first is (in my notaton) $A^c$. You cannot simply change the meaning of $E$ (which is an event in experiment $\mathcal E_1$). Perhaps the solution given by @DilipSarwate is close to what you are thinking: Think of the experiment in which either $E$ or $F$ occur for the first time. What is the probability that the event that occurred was $E$. Jun 22, 2012 at 3:44

I think extreme simplification is need...

$P(E) and P(F)$ are complements for the Universe (U, U=1 in this case) $P(E) + P(F) = 1$ // corrected as mentioned by Aditya, sorry for my dyslexic!thing

$P(E) / ( P(E)+P(F) ) = 1 / 2$ Hence probability of $E$ is $50\%$ (or $0.5$), probability of restant set is the remaining $50\%$; the remaining set is $F$ because $U=\{E, F\}$ with the given data $P(E \text{ before } F) = P(F \text{ before }E)$.

• If $P(E) = P(F) = 1$, then $E$ and $F$ cannot be mutually exclusive because $E \cup F \subset \Omega$, thus $P(E \cup F) = P(E) + P(F) \le P(\Omega) = 1$. Jun 22, 2012 at 3:15
• E and F are exclusive... when I say P(E) = P(F) = 1, I mean "wheight_of_P(E)" + "wheight_of_P(F)" is equal to 1(totality in probability)... without more information from the problem I assume that "E_is_complement_and_exclusive_of_F" and "F_is_complement_and_exclusive_of_E"... I've added parenteses to the answer for clarity...
– ZEE
Jun 22, 2012 at 22:00
• Then you should assume $P(E) = P(F) = 0.5$ Jun 23, 2012 at 1:01
• You're right, what I wanted to say is : P(E) = P(F) and P(E) + P(F) = 1... thanks seeing it... As per opposition to the other possibility which was : P(E) <> P(F) and P(E) + P(F) = 1 in both cases : $P(E) \cap P(F) = \emptyset$ and $P(E) \cup P(F) = U$ (U=Universe or FullSet, 1 in this case)
– ZEE
Jun 23, 2012 at 12:39

Here is an alternative way of using conditional probability.

Consider repeated experiments and let $Z_n$ ($n \in \mathbb{N}$) be the result observed on the $n$-th experiment. Denote the event of "$\textrm{E before F}$" by $B$ and its probability $\alpha$.

According to the law of total probability, we obtain

$$\alpha = P \{ B\} = \sum_{z} P \{B \mid Z_1 = z \} P \{ Z_1 = z \}$$

Now we have

$$P \{ B \mid Z_1 = E \} = 1, \quad P \{ B \mid Z_1 = F \} = 0.$$

If the first experiment results in anything other than $E$ or $F$, the problem is repeated in a statistically identical setting. That is,

$$P \{ B \mid Z_1 = z \} = \alpha, \forall z \neq E, F.$$

Therefore, we have

$$\alpha = P \{ Z_1 = E \} \times 1 + P \{ Z_1 = F \} \times 0 + \sum_{z \neq E,F} P \{ Z_1 = z \} \times \alpha \\ = P \{ Z_1 = E \} + [1 - P \{ Z_1 = E \} - P \{ Z_1 = F \}] \alpha$$

Solving this equation, we get

$$\alpha = \frac{P \{ Z_1 = E \}}{P \{ Z_1 = E \} + P \{ Z_1 = F \}}.$$