Problem 2 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Two cards are drawn successively with replacement from well-shuffled pack of 52 cards. Find the probability distribution of the number of aces. [CBSE-1995, 2001]

Step-by-Step Solution

Answer

The probability distribution is: \( P(X=0) = \left(\frac{12}{13}\right)^2 \), \( P(X=1) = 2 \cdot \frac{1}{13} \cdot \frac{12}{13} \), \( P(X=2) = \left(\frac{1}{13}\right)^2 \).

Step 1: Understanding the Problem

We are working with a standard deck of 52 cards and want to determine the probability distribution for obtaining aces in two draws with replacement. Since there are 4 aces out of 52 cards, the probability of drawing an ace on a single draw is \( \frac{4}{52} = \frac{1}{13} \).

Step 2: Define the Random Variable

Let \( X \) be the random variable representing the number of aces drawn. Since there are two draws and each draw can either result in an ace or not, \( X \) can take values 0, 1, or 2.

Step 3: Calculate Probability of Zero Aces

If no ace is drawn in both draws, the probability for each draw not being an ace is \( \frac{12}{13} \). Therefore, the probability of drawing zero aces is \( P(X=0) = \left(\frac{12}{13}\right)^2 \).

Step 4: Calculate Probability of One Ace

For exactly one ace, it can occur in one of two ways: drawing an ace first followed by a non-ace, or vice versa. The probability for each of these scenarios is \( \frac{1}{13} \cdot \frac{12}{13} \). Since there are two such scenarios, \( P(X=1) = 2 \cdot \frac{1}{13} \cdot \frac{12}{13} \).

Step 5: Calculate Probability of Two Aces

Both draws resulting in an ace means \( P(X=2) = \left(\frac{1}{13}\right)^2 \).

Step 6: Summarize the Probability Distribution

The resulting probability distribution is: - \( P(X=0) = \left(\frac{12}{13}\right)^2 \)- \( P(X=1) = 2 \cdot \frac{1}{13} \cdot \frac{12}{13} \)- \( P(X=2) = \left(\frac{1}{13}\right)^2 \)

Key Concepts

Random Variables

In probability theory, a random variable is a fundamental concept used to describe potential outcomes of a statistical experiment. It is a variable that can take on different values based on the random factors affecting the experiment.
In simpler terms, if you think of each random experiment or trial as a game of chance, the random variable is the aspect you're interested in measuring or observing during the game.
For the given exercise, we defined the random variable \( X \) to represent the number of aces drawn in two draws from a deck of cards. Since each draw can result in either an ace or not, \( X \) can possibly be 0 (no ace), 1 (one ace), or 2 (two aces).
By setting \( X \) with these possible values, it allows us to calculate the likelihood—or probability—of each outcome.

Probability Theory

Probability theory is the branch of mathematics concerned with analyzing random phenomena. It provides a framework for quantifying uncertainty and making predictions about future events based on mathematical principles.
In the context of the given exercise, probability theory helps us calculate the chance of drawing a certain number of aces in two card draws with replacement.

• The probability of drawing an ace is \( \frac{1}{13} \) because there are 4 aces in a 52-card deck.
• Since the cards are drawn with replacement, the probability remains the same for each draw.
• This setup allows us to calculate each possible outcome and compile a probability distribution for the number of aces drawn.

By applying probability theory, we can logically determine the chances for each possible outcome, allowing for informed predictions and decision-making.

Card Probability

Card probability specifically refers to calculating the chances of certain outcomes occurring when drawing cards from a standard deck. This involves understanding the composition of a deck and how different draws affect probability.
In a standard deck, there are 52 cards, of which 4 are aces (one for each suit).
Each card draw is an independent event when drawing with replacement, meaning the result of one draw does not impact the other.

• To find the probability of drawing 0 aces in two draws: for each draw, the probability of not getting an ace is \( \frac{12}{13} \), and for both, it is \( \left(\frac{12}{13}\right)^2 \).
• To find the probability of drawing exactly 1 ace, the ace can appear in the first or second draw: \( 2 \times \frac{1}{13} \times \frac{12}{13} \).
• For 2 aces in both draws, the probability remains \( \left(\frac{1}{13}\right)^2 \).

By systematically evaluating each scenario, card probability helps to clearly understand outcomes and their likelihood in card games or similar activities.