Cards are drawn one-by-one at random from, a well shuffled full pack of 52 playing cards until 2 aces are obtained for the first time . If N is the number of cards required to be drawn. Then `P(N=n)` (where `2 le n le 50`) is given by.

Options

A. `((n-1)(52-n)(51-n))/(50.49.17.13)`
B. `(.^(4)C_(1).^(48)C_(n-2))/(.^(52)C_(n-1)).(3)/(53-n)`.
C. `((48)!)/((n-2)(50-n)!)`
D. none of these

Answer

To solve the problem of finding \( P(N = n) \) where \( N \) is the number of cards drawn until the first occurrence of 2 aces, we can follow these steps:

### Step 1: Understanding the Problem
We need to find the probability that the \( n \)-th card drawn is the second ace, and exactly one ace is drawn among the first \( n-1 \) cards.

### Step 2: Setting Up the Scenario
1. The total number of cards is 52.
2. There are 4 aces in the deck.
3. We need to draw \( n \) cards such that:
- The \( n \)-th card is the second ace.
- Among the first \( n-1 \) cards, there is exactly one ace.

### Step 3: Counting the Ways
To find \( P(N = n) \):
1. Choose 1 ace from the 4 available aces. This can be done in \( \binom{4}{1} \) ways.
2. Choose \( n-2 \) non-ace cards from the remaining 48 cards (since we have already chosen 1 ace). This can be done in \( \binom{48}{n-2} \) ways.
3. The \( n-1 \) cards consist of 1 ace and \( n-2 \) non-aces, and the \( n \)-th card must be an ace. Thus, we can arrange these \( n-1 \) cards in \( \frac{(n-1)!}{(n-2)!} = n-1 \) ways.

### Step 4: Total Combinations
The total number of ways to choose and arrange these cards is:
\[ \text{Total Ways} = \binom{4}{1} \cdot \binom{48}{n-2} \cdot (n-1) \]

### Step 5: Total Possible Outcomes
The total number of ways to choose any \( n-1 \) cards from 52 cards is:
\[ \binom{52}{n-1} \]
The \( n \)-th card can be any of the remaining cards, which is \( 52 - (n-1) = 53 - n \).

### Step 6: Final Probability Calculation
Thus, the probability \( P(N = n) \) is given by:
\[ P(N = n) = \frac{\binom{4}{1} \cdot \binom{48}{n-2} \cdot (n-1)}{\binom{52}{n-1} \cdot (53-n)} \]

### Step 7: Simplifying the Expression
Substituting the values of combinations:
\[ P(N = n) = \frac{4 \cdot \binom{48}{n-2} \cdot (n-1)}{\frac{52!}{(n-1)!(53-n)!} \cdot (53-n)} \]
This simplifies to:
\[ P(N = n) = \frac{4(n-1) \cdot \binom{48}{n-2}}{\binom{52}{n-1} \cdot (53-n)} \]

### Final Answer
Thus, the final expression for \( P(N = n) \) is:
\[ P(N = n) = \frac{4(n-1) \cdot \binom{48}{n-2}}{\binom{52}{n-1} \cdot (53-n)} \]