1. Define events

Let:

• $M_S$ = the missing card is a spade
• $M_N$ = the missing card is not a spade
• $E$ = when two cards are drawn from the remaining pack, both are spades

We need to find: $P(M_N\mid E)$

Using Bayes' theorem, $P(M_N\mid E)=\frac{P(E\mid M_N)P(M_N)}{P(E\mid M_N)P(M_N)+P(E\mid M_S)P(M_S)}$

2. Find prior probabilities

A standard deck has $52$ cards, with:

• $13$ spades
• $39$ non-spades

Since one card is missing at random: $P(M_S)=\frac{13}{52}=\frac14$ $P(M_N)=\frac{39}{52}=\frac34$

3. Find conditional probabilities

Case 1: Missing card is not a spade

Then all $13$ spades are still present among the remaining $51$ cards.

So, $P(E\mid M_N)=\frac{\binom{13}{2}}{\binom{51}{2}}$

Case 2: Missing card is a spade

Then only $12$ spades remain among the $51$ cards.

So, $P(E\mid M_S)=\frac{\binom{12}{2}}{\binom{51}{2}}$

4. Apply Bayes' theorem

$P(M_N\mid E)=\frac{\frac{\binom{13}{2}}{\binom{51}{2}}\cdot \frac34}{\frac{\binom{13}{2}}{\binom{51}{2}}\cdot \frac34+\frac{\binom{12}{2}}{\binom{51}{2}}\cdot \frac14}$

Cancel $\binom{51}{2}$ from numerator and denominator:

$P(M_N\mid E)=\frac{\binom{13}{2}\cdot 3}{\binom{13}{2}\cdot 3+\binom{12}{2}}$

Now, $\binom{13}{2}=78, \qquad \binom{12}{2}=66$

Thus, $P(M_N\mid E)=\frac{78\cdot 3}{78\cdot 3+66}$

$=\frac{234}{234+66}$

$=\frac{234}{300}$

$=\frac{39}{50}$

5. Check options

The required probability is: $\boxed{\frac{39}{50}}$

This matches Option A.

6. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So, they agree.