This problem requires a careful application of Bayes' Theorem, which allows us to update the probability of an event based on new evidence. Here, we are initially uncertain about a missing card, and then we gain new information by drawing two spades. We want to use this new information to revise our belief about the missing card's suit.

1. Key Concepts and Formulas

   • Bayes' Theorem: For two events $A$ and $B$, the conditional probability of $A$ given $B$ is: $P(A|B) = \frac{P(B|A) P(A)}{P(B)}$
   • Law of Total Probability: If $A_1, A_2, \dots, A_n$ form a partition of the sample space (i.e., they are mutually exclusive and exhaustive), then the probability of an event $B$ is: $P(B) = \sum_{i=1}^{n} P(B|A_i)P(A_i)$
   • Combinations: The number of ways to choose $k$ items from a set of $n$ items is given by $C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}$. This is used to calculate probabilities of drawing specific cards.

2. Step-by-Step Solution

   Step 1: Define events and their prior probabilities. A standard deck of 52 cards has 13 spades (S) and 39 non-spades (NS). One card is missing.

   • Let $M_S$ be the event that the missing card is a spade.

   • Let $M_{NS}$ be the event that the missing card is not a spade.

   • Let $E$ be the event that two cards drawn randomly from the remaining 51 cards are found to be spades.

   • Why: We first establish the initial probabilities of the missing card being a spade or not a spade. These are our prior probabilities before any new information (drawing cards) is considered.

   • Calculation: Since any of the 52 cards could be missing with equal probability: $P(M_S) = \frac{\text{Number of spades}}{\text{Total cards}} = \frac{13}{52} = \frac{1}{4}$ $P(M_{NS}) = \frac{\text{Number of non-spades}}{\text{Total cards}} = \frac{39}{52} = \frac{3}{4}$

   Step 2: Calculate the conditional probability of drawing two spades given the missing card is a spade, $P(E | M_S)$.

   • Why: We need to determine the likelihood of observing our evidence (drawing two spades) under the hypothesis that the missing card is a spade.

   • Calculation: If the missing card is a spade ($M_S$), then the remaining deck has $52-1 = 51$ cards. The composition of this deck is:

     • Number of spades = $13 - 1 = 12$
     • Number of non-spades = $39$ (unchanged) The probability of drawing 2 spades from these 51 cards is: $P(E | M_S) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{\frac{12 \times 11}{2}}{\frac{51 \times 50}{2}} = \frac{12 \times 11}{51 \times 50} = \frac{132}{2550} = \frac{22}{425}$

   Step 3: Calculate the conditional probability of drawing two spades given the missing card is not a spade, $P(E | M_{NS})$.

   • Why: Similarly, we need the likelihood of observing our evidence (drawing two spades) under the alternative hypothesis that the missing card is not a spade.

   • Calculation: If the missing card is not a spade ($M_{NS}$), then the remaining deck has $52-1 = 51$ cards. The composition of this deck is:

     • Number of spades = $13$ (unchanged)
     • Number of non-spades = $39 - 1 = 38$ The probability of drawing 2 spades from these 51 cards is: $P(E | M_{NS}) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{\frac{13 \times 12}{2}}{\frac{51 \times 50}{2}} = \frac{13 \times 12}{51 \times 50} = \frac{156}{2550} = \frac{26}{425}$

   Step 4: Calculate the total probability of drawing two spades, $P(E)$, using the Law of Total Probability.

   • Why: Bayes' Theorem requires the marginal probability of the evidence, $P(E)$, in its denominator. We calculate this by summing the probabilities of observing $E$ under each possible scenario for the missing card, weighted by their prior probabilities.

   • Calculation: $P(E) = P(E | M_S) P(M_S) + P(E | M_{NS}) P(M_{NS})$ $P(E) = \left(\frac{22}{425}\right) \times \left(\frac{1}{4}\right) + \left(\frac{26}{425}\right) \times \left(\frac{3}{4}\right)$ $P(E) = \frac{22}{1700} + \frac{78}{1700} = \frac{100}{1700} = \frac{1}{17}$

   Step 5: Apply Bayes' Theorem to find the posterior probability $P(M_{NS} | E)$.

   • Why: We now use Bayes' Theorem to find the probability that the missing card is not a spade, given that we have observed two spades drawn. This is the question explicitly asked.

   • Calculation: $P(M_{NS} | E) = \frac{P(E | M_{NS}) P(M_{NS})}{P(E)}$ $P(M_{NS} | E) = \frac{\left(\frac{26}{425}\right) \times \left(\frac{3}{4}\right)}{\frac{1}{17}}$ $P(M_{NS} | E) = \frac{\frac{78}{1700}}{\frac{1}{17}} = \frac{78}{1700} \times 17$ $P(M_{NS} | E) = \frac{78 \times 17}{1700} = \frac{78 \times 17}{100 \times 17} = \frac{78}{100} = \frac{39}{50}$

3. Common Mistakes & Tips

   • Confusing Conditional Probabilities: A common error is to confuse $P(A|B)$ with $P(B|A)$. Always clearly define your events and what you are trying to find.
   • Incorrect Deck Composition: When a card is missing, the total number of cards and the count of specific suits change. Ensure these changes are correctly reflected in your combinations calculations.
   • Overlooking Law of Total Probability: Forgetting to calculate $P(E)$ (the denominator in Bayes' Theorem) by considering all possible prior states (missing card is spade OR not spade) is a frequent mistake.

4. Summary

   We used Bayes' Theorem to find the probability that the missing card is not a spade, given that two drawn cards were spades. We first established the prior probabilities of the missing card's suit. Then, we calculated the likelihood of drawing two spades under each hypothesis (missing spade vs. missing non-spade). Finally, we combined these using the Law of Total Probability to find the overall probability of drawing two spades and applied Bayes' Theorem to get the posterior probability. The calculated probability is $\frac{39}{50}$.

5. Final Answer The final answer is $\boxed{\frac{39}{50}}$ which corresponds to option (A).