Problem 6 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Find the probability of drawing a diamond card in each of the two consecutive draws from a well-shuffled pack of cards, if the card drawn is not replaced after the first draw. [CBSE-2002(C)]

Step-by-Step Solution

Answer

The probability of drawing a diamond in both draws is \( \frac{1}{17} \).

Step 1: Understand the Total Number of Outcomes

A standard deck of cards contains 52 cards. Each card is unique, so the total number of possible outcomes for the first draw is 52.

Step 2: Calculate Probability for First Draw

In a deck of 52 cards, there are 13 diamonds. So, the probability of drawing a diamond on the first draw is \( \frac{13}{52} \).

Step 3: Adjust Deck for Second Draw

Since the card drawn in the first draw is not replaced, there are now 51 cards remaining.

Step 4: Calculate Probability for Second Draw

Assuming the first card drawn was a diamond, there are now 12 diamond cards left in the deck of 51 cards. The probability of drawing another diamond card is \( \frac{12}{51} \).

Step 5: Find Combined Probability for Both Draws

To find the probability of both events happening (drawing a diamond both times), multiply the probabilities of each event: \( \frac{13}{52} \times \frac{12}{51} = \frac{1}{17} \).

Key Concepts

Understanding Conditional Probability

Conditional probability refers to the likelihood of an event occurring given that another event has already occurred. It is expressed mathematically as \( P(A|B) \), where \( P(A|B) \) is the probability of event \( A \) occurring given event \( B \) has occurred. This concept is crucial when the outcome of one event affects the outcome of the second.
In our exercise, once we draw a card from the deck, it changes the set of possible outcomes for the second draw. This is because the card drawn is not replaced, affecting both the total number of cards and the number of diamonds available for the second draw.
To calculate conditional probability, it's essential to understand how one event influences the next. In our example, knowing the first card drawn was a diamond, we reassess probabilities based on this new scenario. Therefore, conditional probability guides us to recalculate chances dynamically as events happen.

Exploring Combinatorics

Combinatorics is the branch of mathematics dealing with combinations, permutations, and the counting of outcomes. It is often used to make sense of probability-related questions by systematically counting possibilities.
In a card-drawing scenario, combinatorics helps us figure out how many ways we can achieve a specific outcome. By understanding that there are 13 possible diamond cards in a deck of 52, combinatorics provides the foundation to calculate the initial probabilities. Let's consider the adjustments needed when dealing with multiple draws as seen in conditional probability. After one diamond is drawn, combinatorics helps us understand that there are only 12 diamond cards left among 51 cards in the deck, recalculating possibilities for the second draw. Thus, through careful counting and consideration of events, combinatorics becomes an essential tool in complex probability calculations.

Solving Card Probability Problems

Card probability problems often involve determining the chances of drawing specific cards from a deck. These problems integrate both conditional probability and combinatorics to provide complete and accurate solutions.
A standard deck has 52 cards divided into four suits – clubs, diamonds, hearts, and spades. Each suit consists of 13 cards, which gives our exercise its basis: calculating the odds of pulling a diamond card twice in succession, without replacement.
When you draw the first card, there is a straightforward probability of selecting a diamond:

• 13 diamonds out of 52 cards, which simplifies to \( \frac{1}{4} \).

With the card removed, the second draw's probability depends on the drawn card:

• If a diamond, the second draw is \( \frac{12}{51} \), since there are 12 diamonds but only 51 cards left.

Multiplying these probabilities, \( \frac{13}{52} \times \frac{12}{51} \), gives the final result: \( \frac{1}{17} \). Understanding and clearly defining these steps is key to mastering card probability problems.