Abe is in a group with three other people. Each person randomly selects a card from a deck of 52 cards and returns it to the deck. Using permutations, which of the following correctly shows how to find the probability of no two people selecting the same card?

Question

A.

B.

C.

D.

Answer 1

To solve the problem of finding the probability that no two people in a group of four select the same card when each person randomly selects from a standard deck of 52 cards, we can use permutations.

1. The total number of ways for each of the four people to select a card from the deck is \(52^4\). This is because each of the four people can choose from all 52 cards, and they return the card to the deck after selecting.

2. To find the number of favorable outcomes where no two people choose the same card, we can use permutations:

   • The first person can choose any of the 52 cards.
   • The second person can then choose from the remaining 51 cards.
   • The third person can choose from 50 cards.
   • The fourth person can choose from 49 cards.
   • So, the number of favorable outcomes (where no two people select the same card) is given by: \[ 52 \times 51 \times 50 \times 49 \]

3. Therefore, the probability \(P\) that no two people select the same card is: \[ P = \frac{52 \times 51 \times 50 \times 49}{52^4} \]

This can be expressed in terms of permutations as well. The notation for this calculation can be:

\[ P = \frac{P(52, 4)}{52^4} \]

Where \(P(52, 4)\) represents the number of permutations of 52 items taken 4 at a time (which equals \(52 \times 51 \times 50 \times 49\)).

So, to answer the question, we would look for the option that matches the reasoning above.

If the options (A, B, C, D) provided in your original query have different formulas or descriptions, you will have to match the calculation of \(P\) as described with one of those options.