Finding probability of a practical test from independent events

Expert intro: This is a two-event probability problem that combines independence with the probability of exactly one success. The main task is to translate the wording into algebra and solve for the unknown practical-test probability.

Detailed walkthrough

Translate the information into probabilities

Let $T$ be the event of passing the theory test and $P$ be the event of passing the practical test.

We are given:

• $P(T)=0.75$
• The probability of passing only one of the two parts is $0.36$
• The events are independent

If the events are independent, then

$P(T\cap P)=P(T)P(P)=0.75p,$

where $p=P(P)$.

Use the “only one” condition

Passing only one part means either:

• pass theory and fail practical, or
• fail theory and pass practical.

So

$P(T\cap P^c)+P(T^c\cap P)=0.36.$

Using independence:

$P(T\cap P^c)=P(T)P(P^c)=0.75(1-p),$

and

$P(T^c\cap P)=P(T^c)P(P)=0.25p.$

Therefore,

$0.75(1-p)+0.25p=0.36.$

Solve for the practical-test probability

Expand and simplify:

$0.75-0.75p+0.25p=0.36$ $0.75-0.50p=0.36$ $0.50p=0.39$ $p=0.78.$

So the probability of passing the practical test is

$\boxed{0.78}.$

Quick check

If $P(P)=0.78$, then:

• pass theory only: $0.75(0.22)=0.165$
• pass practical only: $0.25(0.78)=0.195$
• total only one: $0.165+0.195=0.36$

This matches the condition, so the answer is consistent.

💡 Pitfall guide

The most common trap is treating “passing only one of the two parts” as if it meant exactly one part out of two without writing both cases. You must include both theory-pass/practical-fail and theory-fail/practical-pass. Another mistake is forgetting independence and writing $P(T\cap P)=P(T)+P(P)$ or assuming the two events are mutually exclusive. Independent events are not mutually exclusive; in fact, they can happen together, and that overlap is essential for setting up the equation correctly.

🔄 Real-world variant

If the question instead said the probability of passing both parts is $0.36$, the setup would change completely. You would write $0.75p=0.36$, giving $p=0.48$. If the events were not independent, the information given here would not be enough to determine $P(P)$ uniquely, because you would need an additional relationship such as a conditional probability or a joint probability table.

🔍 Related terms

independent events, complement probability, exactly one event