Question

From a pack of 52 playing cards, one card was found missing. From the remaining cards, two cards are drawn at random and found to be spade cards. The probability that the missing card is a spade card is

• A. \( \frac{39}{50} \)
• B. \( \frac{27}{51} \)
• C. \( \frac{11}{50} \)
• D. \( \frac{11}{100} \)

Hint:

When working with Bayes' Theorem expansions, avoid multiplying out large product components too early. Leave them in factored form, as huge chunks of numbers will almost always cancel out nicely in the final fractional step!

Answer 1

The Correct Option is C

Concept: This problem can be elegantly solved using Bayes' Theorem. Let us define the two initial mutual hypotheses:

• \( E_1 \): The missing card is a spade. \( P(E_1) = \frac{13}{52} = \frac{1}{4} \)

• \( E_2 \): The missing card is not a spade. \( P(E_2) = \frac{39}{52} = \frac{3}{4} \)
Let \( A \) be the event of drawing 2 spade cards from the remaining 51 cards.

Step 1: Calculating conditional probabilities for event A.

• Given \( E_1 \) (a spade is missing), there are 12 spades left out of 51 cards: \[ P(A|E_1) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{12 \times 11}{51 \times 50} \]

• Given \( E_2 \) (a non-spade is missing), there are 13 spades left out of 51 cards: \[ P(A|E_2) = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{13 \times 12}{51 \times 50} \]

Step 2: Applying Bayes' Theorem.
We want to find \( P(E_1|A) \): \[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \] We can cancel out the common denominator \( \binom{51}{2} \) from all terms: \[ P(E_1|A) = \frac{\frac{1}{4} \times (12 \times 11)}{\left(\frac{1}{4} \times 12 \times 11\right) + \left(\frac{3}{4} \times 13 \times 12\right)} \] We can also cancel out the common factor \( \frac{1}{4} \times 12 \) from the numerator and denominator: \[ P(E_1|A) = \frac{11}{11 + (3 \times 13)} = \frac{11}{11 + 39} = \frac{11}{50} \] This matches option (C) perfectly.