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Chapter 8: Problem 456

If 4 cards are drawn at random and without replacement from a deck of 52 playing cards, what is the chance of drawing the 4 aces as the first 4 cards?

Short Answer

Expert verified
The probability of drawing the 4 aces as the first 4 cards is \(\frac{1}{270,725}\).

Step by step solution

01

Calculate the total number of possible outcomes.

Since there are 52 cards in the deck and we are drawing 4 cards, the total number of possible outcomes can be calculated using the combination formula: \[ C(n,r) = \frac{n!}{r!(n-r)!} \] where \(n\) is the total number of elements and \(r\) is the number of elements to be chosen. For this problem, \(n = 52\) and \(r = 4\). \[ C(52,4) = \frac{52!}{4!(52-4)!} = \frac{52!}{4!48!} \]
02

Calculate the favorable outcomes.

The favorable outcome for this problem is drawing 4 aces as the first 4 cards. Since there are 4 aces in the deck, and we want to pick all 4 aces, the number of favorable outcomes can be calculated using the combination formula: \[ C(4,4) = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} \] Since \(0! = 1\), the result is: \[ C(4,4) = \frac{4!}{4!1} = 1 \]
03

Calculate the probability.

The probability of drawing 4 aces as the first 4 cards can be calculated using the formula: \[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} \] Substituting the values we calculated in steps 1 and 2: \[ P = \frac{1}{C(52,4)} = \frac{1}{\frac{52!}{4!48!}} \]
04

Simplify the result.

Let's simplify the fraction: \[ P = \frac{4!48!}{52!} \] \[ P = \frac{4!}{(49)(50)(51)(52)} \] The factorial \(4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\). Now, we can further simplify the fraction: \[ P = \frac{24}{49 \cdot 50 \cdot 51 \cdot 52} \]
05

Calculate the final probability.

Multiplying and dividing the numerator and denominator, we get: \[ P = \frac{24}{6,497,400} \] \[ P = \frac{1}{270,725} \] So, the probability of drawing the 4 aces as the first 4 cards is \(\frac{1}{270,725}\).

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