Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denot

Q: Why do we multiply by 2 for $P(X=1)$? Can't we just do $p \times q$?

Great question! When we say $X=1$, we mean exactly one Ace. This can happen as (Ace, Non-Ace) OR (Non-Ace, Ace). Since these are mutually exclusive events, we add their probabilities. $p \times q + q \times p = 2pq$.

Q: Why can't we use the hypergeometric distribution here?

Hypergeometric is for sampling *without* replacement. Because we replace the card, the probability remains constant, making it a Binomial experiment.

Q: Is there a faster way to solve this?

Yes! You could calculate $1 - P(X=0)$. Since $P(X=0) + P(X=1) + P(X=2) = 1$, finding $P(X=0)$ (no aces) is often faster. $P(X=0) = q^2 = (\frac{12}{13})^2 = \frac{144}{169}$. Then $1 - \frac{144}{169} = \frac{25}{169}$.

Question

Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then $P(X = 1) + P(X = 2)$ equals :

Options:

(A) 52/169

(B) 25/169

(C) 49/169

(D) 24/169

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITian. Today, we are going to peel back the layers of a classic probability problem. Imagine you are sitting at a table with a standard, well-shuffled deck of $52$ cards.

We are drawing two cards, one after the other, with a crucial twist: **with replacement**.

### The Power of Independence

What does 'with replacement' actually mean for us? It means that after we draw the first card and observe it, we put it back. The deck is restored to its original state.

This is the secret key to the problem. Because the deck is reset, the second draw is completely independent of the first.

The probability of drawing an Ace on the second attempt is exactly the same as it was on the first. This is the beauty of **independent events**.

### Defining Success and Failure

Let us define our success as drawing an Ace. There are $4$ Aces in a deck of $52$ cards.

The probability of drawing an Ace in a single trial, denoted by $p$, is:

$$p = \frac{4}{52} = \frac{1}{13}$$

Conversely, the probability of not drawing an Ace, denoted by $q$, is:

$$q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$$

### The Anatomy of $X$

The random variable $X$ represents the number of Aces we get in two draws. We want to find $P(X=1) + P(X=2)$.

For $X=1$, we need exactly one Ace. This can happen in two distinct ways: (Ace, Non-Ace) or (Non-Ace, Ace).

Mathematically, this is:

$$P(X=1) = (p 	imes q) + (q 	imes p) = 2pq$$

Substituting our values:

$$P(X=1) = 2 	imes \left( \frac{1}{13} ight) 	imes \left( \frac{12}{13} ight) = \frac{24}{169}$$

Now, for $X=2$, we need an Ace in both draws. There is only one way this happens: (Ace, Ace).

$$P(X=2) = p 	imes p = p^2 = \left( \frac{1}{13} ight)^2 = \frac{1}{169}$$

### The Final Synthesis

We are almost there. The question asks for the sum of these probabilities:

$$P(X=1) + P(X=2) = \frac{24}{169} + \frac{1}{169} = \frac{25}{169}$$

It is elegant, is it not? We started with a deck of cards and ended with a precise fraction.

Remember, in probability, the most important step is to clearly define your events and understand the nature of your trials. The final answer is **$\frac{25}{169}$**.

Animated Solution for Mathematics - Probability: Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then $P (X = 1) + P (X = 2)$ equals :

Select Answer:

Visualized Solution

The Experiment Setup

With Replacement

Probability of an Ace ( $p$ )

Simplifying $p$

Probability of a Non-Ace ( $q$ )

Simplifying $q$

The Random Variable $X$

Scenario for $X = 1$

Calculating $P (X = 1)$

Value of $P (X = 1)$

Scenario for $X = 2$

Value of $P (X = 2)$

Final Summation

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Power of Independence

Defining Success and Failure

The Anatomy of $X$

The Final Synthesis

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Visualized Solution

The Experiment Setup

With Replacement

Probability of an Ace ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Probability of a Non-Ace ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q)

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

The Random Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Scenario for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=1

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=1)

Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(X=1)

Scenario for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X=2

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