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 '12 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 '12 at 2:43
  • 1
    $\begingroup$ No, that is a separate issue. In fact, there is no need to assume that $E$ and $F$ are exhaustive. $\endgroup$ – user940 Jun 22 '12 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 $\mathcal E_1$.

  • $\begingroup$ 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 '12 at 3:02
  • 1
    $\begingroup$ $E^c = \{3,4,5,6\} \not\equiv \{3,4\} = F$ $\endgroup$ – Aditya Jun 22 '12 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 '12 at 3:20
  • 1
    $\begingroup$ Does my updated answer clarify this point? $\endgroup$ – Aditya Jun 22 '12 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 '12 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 '12 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 '12 at 22:00
  • $\begingroup$ Then you should assume $P(E) = P(F) = 0.5$ $\endgroup$ – Aditya Jun 23 '12 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 '12 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.