Problem 3 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

We draw two cards from a regular deck of \(52 .\) Let \(S_{1}\) be the event "the first one is a spade," and \(S_{2}\) "the second one is a spade." a. Compute \(\mathrm{P}\left(S_{1}\right), \mathrm{P}\left(S_{2} \mid S_{1}\right)\), and \(\mathrm{P}\left(S_{2} \mid S_{1}^{c}\right)\). b. Compute \(\mathrm{P}\left(S_{2}\right)\) by conditioning on whether the first card is a spade.

Step-by-Step Solution

Answer

\( \mathrm{P}(S_1) = \frac{1}{4}; \mathrm{P}(S_2 \mid S_1) = \frac{12}{51}; \mathrm{P}(S_2 \mid S_1^c) = \frac{13}{51}; \mathrm{P}(S_2) = \frac{1}{4} \).

Step 1: Calculate Probability of First Card Being a Spade

There are 13 spades in a deck of 52 cards. Therefore, the probability of drawing a spade first is given by:\[ \mathrm{P}(S_1) = \frac{13}{52} = \frac{1}{4} \]

Step 2: Conditional Probability of Second Card Being a Spade Given First is a Spade

Once the first card drawn is a spade, there are 12 spades remaining out of 51 cards. Thus, the probability that the second card is a spade given the first was a spade is:\[ \mathrm{P}(S_2 \mid S_1) = \frac{12}{51} \]

Step 3: Conditional Probability of Second Card Being a Spade Given First is Not a Spade

If the first card is not a spade, then all 13 spades are still in the deck of remaining 51 cards. Hence, the probability that the second card is a spade given the first was not a spade is:\[ \mathrm{P}(S_2 \mid S_1^c) = \frac{13}{51} \]

Step 4: Total Probability of Second Card Being a Spade by Conditioning on First Card

To find \( \mathrm{P}(S_2) \), use the law of total probability, considering both scenarios - whether the first card is a spade or not:\[\mathrm{P}(S_2) = \mathrm{P}(S_2 \mid S_1) \cdot \mathrm{P}(S_1) + \mathrm{P}(S_2 \mid S_1^c) \cdot \mathrm{P}(S_1^c) \]Substituting the known probabilities:\[\mathrm{P}(S_2) = \left(\frac{12}{51}\right) \cdot \left(\frac{1}{4}\right) + \left(\frac{13}{51}\right) \cdot \left(\frac{3}{4}\right) \]Simplifying this:\[\mathrm{P}(S_2) = \frac{12}{204} + \frac{39}{204} = \frac{51}{204} = \frac{1}{4}\]

Key Concepts

Conditional Probability

Conditional probability considers the likelihood of an event occurring given that another event has already occurred. It is denoted by \( \text{P}(B \mid A) \), which reads as "the probability of B given A." This condition changes the sample space based on the prior occurrence of event A.
Consider our exercise where the events involve drawing cards from a deck. If you know the first card drawn is a spade, this affects the likelihood of the second card also being a spade.- For example, \( \text{P}(S_2 \mid S_1) = \frac{12}{51} \) means that once the first card is a spade (reducing the deck by one spade), there are now 12 spades left out of 51 cards total.
- Similarly, \( \text{P}(S_2 \mid S_1^c) = \frac{13}{51} \), since if the first isn't a spade, all 13 spades remain in a 51-card deck.
Conditional probability is crucial for understanding relationships between successive events in probability theory, especially when events can influence each other.

Law of Total Probability

The Law of Total Probability aids in finding the probability of an event that depends on several possible conditions or scenarios. It breaks complex problems into simpler, more manageable parts, and combines their probabilities.
In card problems like our exercise, it helps determine the probability of the second card being a spade regardless of the first card's suit.- This law uses: - \( \text{P}(S_2 \mid S_1) \) and the chance of first card being a spade, \( \text{P}(S_1) \) - \( \text{P}(S_2 \mid S_1^c) \) with the chance of the first not being a spade, \( \text{P}(S_1^c) \)
The formula used is:\[\text{P}(S_2) = \text{P}(S_2 \mid S_1) \cdot \text{P}(S_1) + \text{P}(S_2 \mid S_1^c) \cdot \text{P}(S_1^c)\]
In our case, substituting the values leads to \( \text{P}(S_2) = \frac{1}{4} \), illustrating that despite previous conditions, the eventual probability remains consistent through this mathematical decomposition.

Deck of Cards

Playing cards are a fascinating way to illustrate probability concepts, providing a tangible and relatable example. A standard deck consists of 52 cards, divided into four suits: spades, hearts, diamonds, and clubs, each with 13 cards.
Understanding these divisions is vital for calculating probabilities related to card games: - Each suit contributes an equal fraction (like 13 out of 52) in probability computations.
- As draws occur, compositions of the remaining deck change, influencing future probabilities because of the known card positions and absence of replacement. In our scenario, drawing a card without replacement means the first card changes the total deck, affecting future calculation scenarios (such as showing the shift from 52 to 51 cards on subsequent draws).
Working with a deck of cards thus helps illustrate principles like conditional probability and the law of total probability, making abstract concepts more concrete through familiar, real-world examples.