Figuring out a conditional probability question with an answer that doesn’t make sense

Question

Here is the question:

A contractor is hired for the job of paving streets and roads, in the end if the quality of pavement is approved by quality assessments, the contractor's work is accepted.

We know from past experience that $95$ percent of pavement cases are accepted. If the quality assessment is right $75$ percent of the time (without error) about the quality of pavement and is incorrect the remaining $25$ percent of the time, what is the probability that a well-paved road is approved and accepted?

I tried to solve this question but it seems to me that the probability of constructing a well-paved road isn't constant and can be anything between $0.7$ and $0.8$ therefore the probability that a well-paved road would be accepted is anything between $0.7$ and $0.75$.

What are your opinions on this? My numbers don't seem rational to me.

Answer 1

we know from past experience that 95 percent of pavement cases are accepted. if quality assessment is 75 percent of the time right (without errors) about the quality of pavement and 25 percent of times it judges the quality of the pavement wrongly, then what's the probability that a well-paved road is approved and accepted?

Let $A$ be the event of accepting the pavement, and $W$ the event that the pavement is well-paved.

From the text you may evaluate $\mathsf P(A)=0.95, P(A\mid W)=0.75, P(A\mid W^\complement)=0.25$

You seek $\mathsf P(A\cap W)$ which equals $\mathsf P(A\mid W)~\mathsf P(W)$.

So you just need to determine $\mathsf P(W)$.

Answer 2

I believe @GrahamKemp has misinterpreted the question as his answer leads to a probability greater than $1$ (which clearly isn’t right). To stay consistent with his answer, let's stick with the notation $A$ is the event of accepting the pavement, $W$ is the event that the pavement is well paved

It's certainly correct to say that $P(A) = 0.95$.

Where his answer breaks down is that it assumes $P(A | W) = 0.75$ and $P(A|W^c) = 0.25$. However, my understanding was that it is actually saying $P(W|A) = 0.75$ and $P(W|A^c)=0.25$.

We know that $P(A \cap W)=P(A|W)P(W)=P(W|A)P(A)=0.75 \times 0.95 = 0.7125$

The question asks what the probability that a well paved road is accepted is. This is the same as asking for $P(A|W)$ (as stated in your question).

Therefore, there is still some work to do.

We also know that $ P(W \cap A^c)=0.0125$ by using Bayes' Formula.

We now find another expression involving $P(W)$ in order to eliminate this term.

Using Bayes' Formula again, we get:

$$ P(A^c | W)P(W) = P(W \cap A^c) = 0.0125$$

Therefore, we arrive at:

$$ P(A|W)P(W) + P(A^c |W)P(W) = 0.7125 + 0.0125 = 0.725 $$

$$ \iff $$

$$ P(W) \big{(} P(A|W)+P(A^c |W) \big{)} = P(W) = 0.725$$

Finally, we can now substitute back into our original expression:

$$P(A|W)P(W) = 0.725P(A|W) = 0.7125$$

$$ \implies P(A|W) = 0.983 $$