Probability Calculation: Lost Spade Card Scenario

This solution explains how to calculate the value of $n$ in a probability problem involving a lost card from a standard deck and subsequent drawing of spades.

Understanding the Problem Setup

We start with a standard 52-card deck. One card is lost. From the remaining 51 cards, $n$ cards are drawn and all are found to be spades. We are given a specific probability related to the lost card and need to find $n$.

Key information:

• Initial deck size: 52 cards (13 spades, 39 non-spades).
• One card is lost.
• Remaining deck size: 51 cards.
• $n$ cards are drawn from the 51, and all are spades.
• The probability of the lost card being a spade, given the outcome of drawing $n$ spades, is $P(S|E) = \frac{11}{50}$.

Applying Bayes' Theorem for Conditional Probability

Let $S$ be the event that the lost card was a spade. Let $S'$ be the event that the lost card was not a spade. Let $E$ be the event that $n$ cards drawn from the remaining 51 are all spades.

The prior probabilities are:

• $P(S) = \frac{13}{52} = \frac{1}{4}$
• $P(S') = \frac{39}{52} = \frac{3}{4}$

We are given the posterior probability $P(S|E) = \frac{11}{50}$.

Using Bayes' Theorem: $P(S|E) = \frac{P(E|S) P(S)}{P(E)}$ where $P(E) = P(E|S) P(S) + P(E|S') P(S')$.

Calculating Probabilities of Drawing Spades

We need to find the probability of event $E$ (drawing $n$ spades) under the two conditions (lost card was spade or not spade).

• If the lost card was a spade (Event S):
  • Remaining cards: 51 total, 12 spades, 39 non-spades.
  • The probability of drawing $n$ spades is $P(E|S) = \frac{\binom{12}{n}}{\binom{51}{n}}$.
• If the lost card was not a spade (Event S'):
  • Remaining cards: 51 total, 13 spades, 38 non-spades.
  • The probability of drawing $n$ spades is $P(E|S') = \frac{\binom{13}{n}}{\binom{51}{n}}$.

Solving for n

Substitute these into the Bayes' formula:

$ P(S|E) = \frac{\left( \frac{\binom{12}{n}}{\binom{51}{n}} \right) \times \frac{1}{4}}{\left( \frac{\binom{12}{n}}{\binom{51}{n}} \right) \times \frac{1}{4} + \left( \frac{\binom{13}{n}}{\binom{51}{n}} \right) \times \frac{3}{4}} $

Simplify the equation by multiplying the numerator and denominator by $4 \times \binom{51}{n}$:

$ \frac{11}{50} = \frac{\binom{12}{n}}{\binom{12}{n} + 3 \binom{13}{n}} $

Now, we need to find the value of $n$ that satisfies this equation. Let's test integer values.

Test $n=2$:

• $\binom{12}{2} = \frac{12 \times 11}{2 \times 1} = 66$
• $\binom{13}{2} = \frac{13 \times 12}{2 \times 1} = 78$

Substitute these values into the equation:

$ \text{RHS} = \frac{66}{66 + 3 \times 78} = \frac{66}{66 + 234} = \frac{66}{300} $

Simplify the fraction:

$ \frac{66}{300} = \frac{6 \times 11}{6 \times 50} = \frac{11}{50} $

The calculated value matches the given probability $P(S|E)$.

Conclusion on n

The value $n=2$ satisfies the condition derived from Bayes' theorem and the provided probability.