Rom

\(P[\text{knew answer|correct answer}] = \\ \dfrac{P[\text{correct answer | knew answer}]P[\text{knew answer}]}{P[\text{correct answer}]}\)

\(P[\text{knew answer}] = \dfrac 1 2\\ P[\text{correct answer|knew answer}] = 1\\ P[\text{correct answer}] = \dfrac 1 2 + \dfrac 1 2 \left(\dfrac 1 4 \dfrac 1 3 + \dfrac 3 4 \dfrac 1 4\right) =\dfrac{61}{96}\)

\(P[\text{knew answer|correct answer}] = \dfrac{1 \cdot \dfrac 1 2}{\dfrac{61}{96}} = \dfrac{48}{61}\)

Pick a card any card.  Probability of this is 1.

Now pick a card not duplicate to the first card.  Probability of this is 46/47

Continuing we have

\(p=1 \dfrac{46}{47}\dfrac{44}{46}\dfrac{42}{45} \dfrac{40}{44} = \dfrac{112}{141}\)

\(\text{Select the slot that N and D will occupy}\\ \text{There are 4 of these}\\ \text{Select the ordering of N and D within the slot}\\ \text{There are two arrangements}\\ \text{There are now 5 unique remaining letters}\\ \text{These can be arranged 5! ways in the remaining slots}\\ \text{So there are $4 \cdot 2 \cdot 5! = 960$ valid arrangements}\\ \text{There are a total of $7!=5040$ possible arrangements}\\ p = \dfrac{960}{5040} = \dfrac{4}{21} \)

\(p = \dfrac{\dbinom{4}{2}\dbinom{11}{1}+\dbinom{5}{2}\dbinom{10}{1}+\dbinom{6}{2}\dbinom{9}{1}}{\dbinom{15}{3}}\)

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