Working my way through the following 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:


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 $p$)

  • $\begingroup$ They mean: If neither $E$ or $F$ happens on the first trial, then the game starts over. $\endgroup$
    – user940
    Jun 22, 2012 at 2:34
  • $\begingroup$ So we are able to treat the experiment as if only mutually exclusive events $E$ and $F$ exist and my solutions is valid correct? $\endgroup$
    – rudolph9
    Jun 22, 2012 at 2:43
  • 2
    $\begingroup$ No, that is a separate issue. In fact, there is no need to assume that $E$ and $F$ are exhaustive. $\endgroup$
    – user940
    Jun 22, 2012 at 2:47

4 Answers 4


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 $\mathcal E_1$.

  • $\begingroup$ 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? $\endgroup$
    – rudolph9
    Jun 22, 2012 at 3:02
  • 1
    $\begingroup$ $E^c = \{3,4,5,6\} \not\equiv \{3,4\} = F$ $\endgroup$
    – Aditya
    Jun 22, 2012 at 3:12
  • $\begingroup$ 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? $\endgroup$
    – rudolph9
    Jun 22, 2012 at 3:20
  • 1
    $\begingroup$ Does my updated answer clarify this point? $\endgroup$
    – Aditya
    Jun 22, 2012 at 3:25
  • 1
    $\begingroup$ 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 $E$. $\endgroup$
    – Aditya
    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 $P(E \text{ before } F) = P(F \text{ before }E)$.

  • $\begingroup$ If $P(E) = P(F) = 1$, then $E$ and $F$ cannot be mutually exclusive because $E \cup F \subset \Omega$, thus $P(E \cup F) = P(E) + P(F) \le P(\Omega) = 1$. $\endgroup$
    – Aditya
    Jun 22, 2012 at 3:15
  • $\begingroup$ 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... $\endgroup$
    – ZEE
    Jun 22, 2012 at 22:00
  • $\begingroup$ Then you should assume $P(E) = P(F) = 0.5$ $\endgroup$
    – Aditya
    Jun 23, 2012 at 1:01
  • $\begingroup$ 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) $\endgroup$
    – 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

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


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