Finding a Missing Probability Using Independence

Work Backward from the Probability of Both Events

In Union Probability for Independent Events, you used independence to find the probability that both events occur by multiplying their individual probabilities. Sometimes the probability of both events is known, but one individual event probability is missing. This tutorial uses the same multiplication rule in reverse.

Let $A$ and $B$ be independent events. If $P(A)$ and $P(A\cap B)$ are known, the independent-events multiplication rule says that $P(A\cap B)=P(A)P(B)$. When $P(A)>0$, divide both sides by $P(A)$ to isolate $P(B)$.

P(B)=\frac{P(A\cap B)}{P(A)},\qquad P(A)>0.

This is not a new rule for finding an intersection. It is an algebraic rearrangement of the multiplication rule for independent events. The intersection probability is the probability that both events occur; it is not itself the missing probability.

Formula: If $A$ and $B$ are independent and $P(A)>0$, then $$ P(B)=\frac{P(A\cap B)}{P(A)}. $$ To check a result, multiply $P(A)$ by the calculated $P(B)$. The product should recover the given $P(A\cap B)$.

The answer also needs to be a possible probability: it must be between $0$ and $1$, inclusive. For example, if the division gives $1.2$, the stated probabilities cannot describe independent events as claimed. The inputs may be inconsistent, or an assumption may have been misunderstood.

Conditions and a Useful Boundary Case

Before dividing, check that the known probability $P(A)$ is positive. Division by zero is not defined. If $P(A)=0$, then $A$ cannot occur, so $P(A\cap B)=0$ for any event $B$. Knowing that the intersection probability is zero in this case does not reveal $P(B)$; $B$ might have any probability from $0$ to $1$.

When $P(A)>0$, a valid intersection probability cannot exceed $P(A)$: the outcomes in $A\cap B$ are part of the outcomes in $A$. Thus, if the given intersection probability is greater than $P(A)$, the inputs are already impossible, even before you calculate the quotient. If $P(A\cap B)$ is between $0$ and $P(A)$, the quotient is between $0$ and $1$.

Conditions: To solve for $P(B)$ from $P(A)$ and $P(A\cap B)$:

The events $A$ and $B$ are independent, so $P(A\cap B)=P(A)P(B)$.
The known probability $P(A)$ is greater than $0$, so division is possible.
The given probabilities are consistent, including $0\leq P(A\cap B)\leq P(A)$.
The calculated value of $P(B)$ must be between $0$ and $1$, inclusive.

Do not confuse this rearrangement with the conditional probability formula from The Conditional Probability Formula. Conditional probability divides the intersection by the probability of the event after the bar. Here, the same-looking quotient $P(A\cap B)/P(A)$ equals $P(B)$ specifically because independence makes $P(A\cap B)=P(A)P(B)$. The independence assumption is what allows the interpretation as the missing marginal probability.

Worked Examples

Worked Example: A Sensor and a Data-Transfer Alert

In a hypothetical equipment model, let $A$ be the event that a temperature sensor activates during a test, and let $B$ be the event that the system sends a data-transfer alert. The model gives $P(A)=0.40$ and $P(A\cap B)=0.18$, and assumes the events are independent. Find $P(B)$.

State. The target is the probability $P(B)$ that the system sends a data-transfer alert during a test.

Plan. The events are stated to be independent, so use $P(A\cap B)=P(A)P(B)$. Since $P(A)=0.40>0$, solve for $P(B)$ by dividing the given intersection probability by $P(A)$. Then check the answer by multiplying back.

Do.

P(B)=\frac{P(A\cap B)}{P(A)} =\frac{0.18}{0.40} =0.45.

The result is between $0$ and $1$. Check it using the multiplication rule: $P(A)P(B)=(0.40)(0.45)=0.18$, which matches the given intersection probability.

Conclude. Under the model’s independence assumption, the probability that the system sends a data-transfer alert during a test is $0.45$, or $45\%$.

Worked Example: A Shuttle and a Charging Station

A hypothetical campus model considers one randomly selected morning. Let $A$ be the event that a shuttle arrives late, and $B$ the event that a particular charging station is unavailable. The model assigns $P(A)=0.72$ and $P(A\cap B)=0.36$, and treats the events as independent. Find the probability that the charging station is unavailable.

State. The missing probability is $P(B)$, the chance that the charging station is unavailable on a randomly selected morning.

Plan. Use the independent-events multiplication rule rearranged to solve for $P(B)$. Confirm that the known probability $P(A)$ is positive and that the resulting probability is valid.

Do. Since $0.72>0$, division is possible. The calculation is

P(B)=\frac{P(A\cap B)}{P(A)} =\frac{0.36}{0.72} =0.50.

The answer is within the probability range. Multiplying back gives $(0.72)(0.50)=0.36$, the stated probability that both events occur.

Conclude. Under the stated independence model, the probability that the charging station is unavailable on a randomly selected morning is $0.50$, or $50\%$.

Worked Example: A Library System Check

A hypothetical library technology model considers one randomly selected day. Let $A$ be the event that the online catalog is temporarily unavailable, and $B$ the event that a self-checkout terminal needs a restart. Suppose $P(A)=0.25$, $P(A\cap B)=0.05$, and the model assumes independence. Find $P(B)$.

State. We want $P(B)$, the probability that a self-checkout terminal needs a restart on a randomly selected day.

Plan. Because the model assumes independence, the intersection is the product of the individual probabilities. Divide the given intersection probability by $P(A)$. Check the result both against the possible probability range and by multiplying back.

Do. The denominator is positive, and $0.05\leq 0.25$, so the supplied values pass the basic input check. Then

P(B)=\frac{P(A\cap B)}{P(A)} =\frac{0.05}{0.25} =0.20.

The result is a possible probability. Checking the product, $(0.25)(0.20)=0.05$, recovers the given intersection probability. As an additional consistency check, $P(A\mid B)=0.05/0.20=0.25=P(A)$, which agrees with the independence condition.

Conclude. Under the hypothetical model, the probability that a self-checkout terminal needs a restart on a randomly selected day is $0.20$, or $20\%$.

When the Numbers Do Not Fit

The calculation can also reveal that the given information is inconsistent with independence. Suppose $P(A)=0.40$ and $P(A\cap B)=0.46$. The intersection cannot have greater probability than event $A$, because every outcome in the intersection is also in $A$. The proposed values therefore cannot be valid event probabilities, regardless of independence.

The quotient makes the problem visible too:

P(B)=\frac{0.46}{0.40}=1.15.

A probability of $1.15$ is impossible. Do not report it as the answer or round it down to $1$. Instead, explain that the information is inconsistent with the stated independence assumption and probability rules. Rounding cannot repair an impossible value.

The zero case is different. If $P(A)=0$, the quotient is undefined, not merely inconvenient. Independence implies $P(A\cap B)=P(A)P(B)=0$ for every possible value of $P(B)$. Therefore, $P(A)=0$ and $P(A\cap B)=0$ do not provide enough information to find $P(B)$.

Common Mistakes and AP Exam Tips

Multiplying when the question asks for the missing event probability. Multiplication finds the intersection from both individual probabilities. Here, rearrange that rule and divide the intersection by the known individual probability.
Forgetting to state independence. The quotient gives $P(B)$ because the events are independent. Without that assumption, $P(A\cap B)/P(A)$ is $P(B\mid A)$, not necessarily $P(B)$.
Dividing by the wrong probability. If the known value is $P(A)$ and the target is $P(B)$, divide $P(A\cap B)$ by $P(A)$. This follows by isolating $P(B)$ in the product $P(A)P(B)$.
Accepting an impossible result. A probability cannot be greater than $1$ or less than $0$. Check the quotient and, before dividing, check that the intersection is no greater than the known event probability.
Trying to divide by zero. If $P(A)=0$, the information does not determine $P(B)$. Explain why rather than attempting the calculation.
Giving only a number. A complete response names the event represented by the answer and describes it in the stated context.

AP Exam Tip: Write the independence equation first, $P(A\cap B)=P(A)P(B)$, then show the algebra that isolates the unknown. State that the known probability is positive, check that the answer is between $0$ and $1$, and verify it by multiplying back. Conclude in context.

Key Takeaway

Independence can be used in either direction. If you know both individual probabilities, multiply to find the chance that both events occur. If you know the probability of both events and one individual probability, divide to find the other individual probability, provided the known probability is positive.

Key takeaway: For independent events with $P(A)>0$, $P(B)=P(A\cap B)/P(A)$. Check that the result is a possible probability and that multiplying it by $P(A)$ recovers the given intersection probability.

Check Your Understanding

For each question, show the calculation and explain what the result means or why the information is insufficient.

Events $A$ and $B$ are independent, $P(A)=0.30$, and $P(A\cap B)=0.12$. Find and interpret $P(B)$.
A model gives independent events with $P(A)=0.80$ and $P(A\cap B)=0.20$. Find $P(B)$, then verify it by multiplication.
For independent events, $P(A)=0.25$ and $P(A\cap B)=0.30$. Explain why the information is inconsistent.
If $P(A)=0$ and $P(A\cap B)=0$, can you determine $P(B)$? Explain.
Explain why $P(A\cap B)/P(A)$ does not always equal $P(B)$ when independence is not established.