An art dealer receives a shipment of five old paintings from abroad, and, on the basis of past experience, she feels that the probabilities are, respectively, 0.76, 0.09, 0.02, 0.01, 0.02, and 0.10 that 0, 1, 2, 3, 4, or all 5 of them are forgeries. Since the cost of authentication is fairly high, she decides to select one of the five paintings at random and send it away for authentication. If it turns out that this painting is a forgery, what probability should she now assign to the possibility that all the other paintings are also forgeries?
So far I tried assigning A as the event all paintings are forgeries and N that none of the paintings are forgeries so the required probability becomes P(A|N*) = $$\left(\frac{P(A \cap N*)}{P(N*)}\right)$$
which gives $\left(\frac{0.10}{0.24}\right)$ cause $P( A \cap N*) = $ P (A) = 0.10 and P(N*) = 1-0.76 = 0.24 and my answer is o.4166 but the textbook gives the answer as 0.6757
Intuitively, the art dealer has more information than just "at least one is a forgery" when she discovers that a particular one is a forgery. This is the reason your answer doesn't match the correct answer.
Let $F_i$ denote the event that $i$ are forgeries and $C$ denote the event that she chooses a forgery. Then
$$\mathbb{P}(F_5 | C) = \frac{\mathbb{P}(F_5)\mathbb{P}(C|F_5)}{\mathbb{P}(C)}$$
A couple of these are trivial: $\mathbb{P}(F_5) = 0.10$ and $\mathbb{P}(C|F_5) = 1$, but the last one requires a little bit of work:
$$\begin{align*}\mathbb{P}(C) &= \sum_{i=0}^5\mathbb{P}(F_i)\mathbb{P}(C|F_i) \\&= \sum_{i=0}^5\mathbb{P}(F_i)\cdot \frac{i}{5} \\&= 0.09\cdot \frac{1}{5}+0.02\cdot\frac{2}{5}+0.01\cdot\frac{3}{5} + 0.02\cdot\frac{4}{5}+0.10 \\&= 0.148\end{align*}$$ (here, we also can note that the probability she chooses a forgery is the expected number of forgeries divided by the number of paintings)
Therefore $$\mathbb{P}(F_5|C) = \frac{0.10}{0.148} \approx 0.675675676$$