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

Step 1: Define the cases.
Let $E_1$ be that the missing card is a spade, with $P(E_1)=\tfrac{13}{52}=\tfrac{1}{4}$, and $E_2$ that it is not, with $P(E_2)=\tfrac{3}{4}$. Let $A$ be drawing 2 spades from the remaining 51.
Step 2: Chance of A if a spade is missing.
Then 12 spades remain: $P(A|E_1) = \dfrac{12\times11}{51\times50}$.
Step 3: Chance of A if no spade is missing.
Then 13 spades remain: $P(A|E_2) = \dfrac{13\times12}{51\times50}$.
Step 4: Write Bayes' formula.
\[ P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)} \]
Step 5: Cancel common parts.
Drop the shared $\tfrac{1}{51\times50}$ and the factor $\tfrac{1}{4}\times12$, leaving $\dfrac{11}{11 + 3\times13}$.
Step 6: Finish the arithmetic.
\[ P = \frac{11}{11+39} = \frac{11}{50} \] \[ \boxed{\tfrac{11}{50}} \]