Problem 36 - A First Course in Probability

Question

Two cards are chosen at random from a deck of 52 playing cards. What is the probability that they (a) are both aces? (b) have the same value?

Step-by-Step Solution

Answer

(a) The probability of both cards being aces is \(\frac{1}{221}\). (b) The probability of both cards having the same value is \(\frac{13}{221}\).

Step 1: Total number of ways to choose two cards from a deck of 52

Using combinations, we can find the total number of ways to choose two cards from a deck of 52, denoted as C(52,2). The formula for combinations is \(C(n,r) = \frac{n!}{r!(n-r)!}\), where n is the total number of items and r is the number of items you are choosing. In our case, n = 52 and r = 2. Total combinations = \(C(52, 2) = \frac{52!}{2!(52-2)!} = \frac{52!}{2!50!} = \frac{52 \times 51}{2} = 1326\) There are 1326 ways to choose two cards from a deck of 52 cards.

Step 2: (a) Both cards are aces

We need to find the probability that both cards are aces. There are 4 aces in the deck, so we first need to find the total number of ways to choose two aces from the 4 available. Using combinations again, we can calculate this as: Total combinations of aces = \(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2} = 6\) There are 6 possible ways to choose two aces from the deck. Now, we can find the probability by dividing the successful outcomes (both cards are aces) by the total possible outcomes: Probability(both aces) = \(\frac{Total \: combinations \: of \: aces}{Total \: combinations} = \frac{6}{1326}\) Probability(both aces) = \(\frac{1}{221}\)

Step 3: (b) Both cards have the same value

Here, we want to find the probability that both cards have the same value (e.g., two 2's, two 3's, ..., two Kings). There are 13 different card values in the deck. For each card value, we have 4 of that card, one in each suit. Let's find the number of ways to choose two cards with the same value for one card value (e.g., two 2's): Combinations for one value = \(C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2} = 6\) There are 6 possible ways to choose two cards with the same value for one card value. We have 13 card values, so we can multiply this by 13 to get the total number of successful outcomes: Total successful outcomes = 6 (combinations for one value) x 13 (card values) = 78 Now, we can find the probability by dividing the successful outcomes (both cards have the same value) by the total possible outcomes: Probability(same value) = \(\frac{Total \: successful \: outcomes}{Total \: combinations} = \frac{78}{1326}\) Probability(same value) = \(\frac{13}{221}\) In conclusion, (a) The probability of both cards being aces is \(\frac{1}{221}\). (b) The probability of both cards having the same value is \(\frac{13}{221}\).