can't get correct answer using different method, conditional probability

Question

A asks B a question, for any possible question B knows good answer $3$ times per $4$ questions but B thinks he always know good answer. B lies one time per $4$ questions. What is the probability B answer corret?

My textbook says $\frac{5}{8}$ and indeed if I list possible outcomes for $4$ questions there are $10$ outcomes that B answer correct (3+3+3+1) and $6$ (1+1+1+3) outcomes that B answer wrong. But when I want to use conditional probability I can't get this answer.

Let C be event that B knows good answer and let D be event that B lies.

$$P(C|D)=\frac{P(C \cap D)}{P(D)}$$

$$\frac{5}{8}=\frac{P(C \cap D)}{\frac{1}{4}}$$

So $P(C \cap D)$ should equal to $\frac{5}{32}$ but I have no idea why.

Answer 1

I think what is meant by the question is that B is given a yes/no question and is correct 3/4 of the time, but answers the opposite of what he thinks 1/4 of the time. If this is the case then the probability of him answering correctly is as below, where the former part represents when he is correct and doesn't lie and the latter where he is wrong and lies (correcting their mistake). $$\frac{3}{4}^2+\frac{1}{4}^2 = \frac{5}{8}$$

The problem with what you have written is that the probability of them replying with the correct answer is not $P(C\mid D)$

Answer 2

As the comment of lulu suggests, in order to attack the problem, you have to assume that in the $(1/16)$-th of the time that Person-B is both mistaken and attempting to lie, that Person-B will then inadvertently answer correctly.

So there are two mutually exclusive ways that Person-B's answer can be correct:

• B knows the right answer, and answers truthfully.
  The probability of this happening is $~\displaystyle \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}.$

• B does not know the right answer, and lies, inadvertantly giving the right answer.
  The probability of this happening is $~\displaystyle \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}.$

So, the combined sum of these two mutually exclusive possibilities is

$$\frac{9}{16} + \frac{1}{16} = \frac{5}{8}.$$

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The real challenge of this problem is trying to reverse engineer the problem composer's intent. Personally, the only way that I can make sense out of the fact that the official answer is $~\frac{5}{8},~$ is by making the assumption shown at the start of my answer.

For what it's worth, I regard the assumption as bizarre and unrealistic, and I suspect that the problem composer is in the wrong line of work.

Note
You also have to assume that the events of B knowing the right answer and B lying are independent events. Given that no information is given to the contrary, and given that this assumption is not that unrealistic, this specific assumption seems reasonable.