Exactly One Event Among Independent Events

What Does “Exactly One” Mean?

When several events are possible, “at least one” and “exactly one” describe different outcomes. Exactly one event occurs when one specified event happens and every other event does not. To calculate its probability, identify each possible event that could be the sole occurrence, calculate the probability of each case, and add those probabilities.

Independence makes the probability of each case a product. For the case in which event $A_i$ alone occurs, $A_i$ must occur while all the other events fail. The result is useful whenever a problem asks for one occurrence among several independent opportunities, rather than one or more occurrences.

Definition: Let $A_1,\ldots,A_n$ be mutually independent events, and write $p_i=P(A_i)$. “Exactly one of the events occurs” is the event that, for some $i$, $A_i$ occurs and every $A_j$ with $j\ne i$ does not. Its probability is $$ P(\text{exactly one})=\sum_{i=1}^{n}p_i\prod_{\substack{j=1\\j\ne i}}^{n}(1-p_j). $$

The product for a particular $i$ represents the case “$A_i$ alone.” For example, with three events, the case in which only $A_2$ occurs has probability $P(A_1^c\cap A_2\cap A_3^c)$. By independence of the events and their complements, this equals $(1-p_1)p_2(1-p_3)$. The cases “only $A_1$,” “only $A_2$,” and so on cannot happen together, so their probabilities are added.

Building and Justifying the Formula

For each $i$, define the event $B_i$ to mean that $A_i$ occurs and all the other $A_j$ do not. The events $B_1,\ldots,B_n$ are mutually exclusive: if $B_i$ occurs, then $A_i$ occurs and every other event fails, so no $B_k$ with $k\ne i$ can also occur. Their union is exactly the event that one of the original events occurs.

Since the original events are mutually independent, probabilities of intersections involving their occurrences and complements factor into products. Thus $P(B_i)=p_i\prod_{j\ne i}(1-p_j)$. The addition rule for mutually exclusive events then gives the formula.

\begin{aligned} P(\text{exactly one}) &=P(B_1\cup\cdots\cup B_n)\\ &=\sum_{i=1}^{n}P(B_i)\\ &=\sum_{i=1}^{n}p_i\prod_{\substack{j=1\\j\ne i}}^{n}(1-p_j). \end{aligned}

The assumption is mutual independence of all $n$ events, not merely that every pair is independent. For two events, pairwise independence is enough because there is only one pair. For three or more events, pairwise independence alone does not in general ensure that intersections involving all the events factor as required. Also, independence must be supplied or established; knowing only the separate probabilities $p_i$ is not enough to calculate the probability of exactly one event.

Useful special case: If all $n$ mutually independent events have the same probability $p$, there are $n$ mutually exclusive ways for exactly one to occur. Each way has probability $p(1-p)^{n-1}$, so $$ P(\text{exactly one})=np(1-p)^{n-1}. $$

Worked Calculations

Worked Example: Three Independent Review Flags

A hypothetical review process has three independent flags. The probabilities that flags $A$, $B$, and $C$ are raised are $0.20$, $0.50$, and $0.10$, respectively. Find the probability that exactly one flag is raised. Assume the flags are mutually independent.

Method. Calculate the three mutually exclusive “only this flag” cases, then add them. The complement probabilities are $P(A^c)=1-0.20=0.80$, $P(B^c)=1-0.50=0.50$, and $P(C^c)=1-0.10=0.90$.

Only $A$. $A$ is raised while $B$ and $C$ are not: $0.20(0.50)(0.90)=0.0900$.

Only $B$. $B$ is raised while $A$ and $C$ are not: $0.80(0.50)(0.90)=0.3600$.

Only $C$. $C$ is raised while $A$ and $B$ are not: $0.80(0.50)(0.10)=0.0400$.

The cases cannot overlap, so add: $0.0900+0.3600+0.0400=0.4900$. The probability that exactly one flag is raised is 0.4900.

Check. The result is between 0 and 1, and each case includes one raised flag and two flags not raised. A common mistake is to add only the flag probabilities, obtaining $0.20+0.50+0.10=0.80$; that counts outcomes with multiple raised flags as well as outcomes with exactly one.

Worked Example: Exactly One Occurrence Among Four Independent Events

In a hypothetical monitoring exercise, four mutually independent events $D_1,D_2,D_3,D_4$ have probabilities $0.10,0.20,0.30,0.40$. Find the probability that exactly one occurs.

Method. Use the general formula by calculating each event’s probability multiplied by the probabilities that the other three events fail. The failure probabilities are $0.90,0.80,0.70,0.60$.

Only $D_1$ occurs: $0.10(0.80)(0.70)(0.60)=0.0336$.

Only $D_2$ occurs: $0.90(0.20)(0.70)(0.60)=0.0756$.

Only $D_3$ occurs: $0.90(0.80)(0.30)(0.60)=0.1296$.

Only $D_4$ occurs: $0.90(0.80)(0.70)(0.40)=0.2016$.

Add the four disjoint case probabilities: $0.0336+0.0756=0.1092$, $0.1092+0.1296=0.2388$, and $0.2388+0.2016=0.4404$. The probability that exactly one event occurs is 0.4404.

Check. The largest single case is “only $D_4$,” consistent with $D_4$ having the largest occurrence probability. A common mistake is to apply $np(1-p)^{n-1}$ using an average probability: that shortcut requires all event probabilities to be equal, which they are not here.

Worked Example: Finding an Unknown Event Probability

Three mutually independent events $A$, $B$, and $C$ have $P(A)=0.20$ and $P(B)=0.40$. The probability that exactly one occurs is $0.46$. Find $P(C)$.

Method. Let $q=P(C)$, so $P(C^c)=1-q$. Write and solve the sum of the three “only this event” cases.

Only $A$ occurs with probability $0.20(0.60)(1-q)=0.12(1-q)$.

Only $B$ occurs with probability $0.80(0.40)(1-q)=0.32(1-q)$.

Only $C$ occurs with probability $0.80(0.60)q=0.48q$.

Set their sum equal to the given probability: $0.12(1-q)+0.32(1-q)+0.48q=0.46$. Combining terms gives $0.44(1-q)+0.48q=0.46$, then $0.44-0.44q+0.48q=0.46$, so $0.04q=0.02$ and $q=0.50$.

Check. Substituting $q=0.50$, the three cases have probabilities $0.06$, $0.16$, and $0.24$; their sum is $0.46$, as required. The unknown probability $0.50$ is valid because it lies between 0 and 1. A common mistake is to treat the given $0.46$ as the probability of $C$; it is the combined probability of all three disjoint cases.

How the Result Differs From “At Least One”

The exactly-one calculation excludes outcomes in which two or more events occur. In contrast, “at least one” includes every outcome with one or more occurrences. For independent events, the probability of at least one is $1-\prod_{i=1}^n(1-p_i)$, because its complement is that all events fail. That complement calculation does not by itself give the probability of exactly one.

For two independent events, the distinction is especially clear. Exactly one means either the first occurs and the second does not, or the second occurs and the first does not:

P(\text{exactly one})=p_1(1-p_2)+(1-p_1)p_2.

By comparison, the probability of at least one is $p_1+p_2-p_1p_2$. The difference between the expressions is the probability $p_1p_2$ that both events occur. In a problem with three or more events, calculate every “only one” case carefully; do not assume that subtracting an “at least one” probability from 1 will isolate exactly one.

Common Exam Traps

Adding the $p_i$ values. This does not require all other events to fail, so it does not generally calculate exactly one.
Forgetting a complement. In the “only $A_i$” case, every event other than $A_i$ must be absent, contributing a factor $1-p_j$.
Using the equal-probability shortcut for unequal events. Use $np(1-p)^{n-1}$ only when all event probabilities are the same and the events are mutually independent.
Assuming independence from the probabilities alone. Separate event probabilities do not determine their joint behavior. Verify that mutual independence is given or justified.
Confusing the cases with independent events. The “only $A_1$” and “only $A_2$” cases are mutually exclusive, so add their probabilities; independence is used to multiply within each case.

Key takeaway: For each possible event that could be the sole occurrence, multiply its probability by the failure probabilities of all the others. Add these mutually exclusive case probabilities, using mutual independence to justify each product.

Check Your Understanding

For each problem, assume the events are mutually independent unless stated otherwise.

Three events have probabilities $0.10$, $0.20$, and $0.30$. Find the probability that exactly one occurs. (Answer: 0.3980)
Four independent events have the same probability $0.20$. Find the probability that exactly one occurs. (Answer: 0.4096)
Two events have probabilities $0.35$ and $0.60$. Calculate the probability that exactly one occurs. (Answer: 0.5300)
Three events have probabilities $0.10$, $0.30$, and $0.50$. Find the probability that exactly one occurs. (Answer: 0.4850)
Events $A$, $B$, and $C$ are mutually independent, with $P(A)=0.25$ and $P(B)=0.50$. If the probability that exactly one occurs is $0.425$, find $P(C)$. (Answer: 0.6000)

What You'll Learn