A standard deck of playing cards has 52 cards divided equally into four suits: hearts , diamonds , clubs , and spades . Within each suit, there are 13 cards: one ace, the numbers from 2 to 10, a jack, queen, and king.Suppose you are dealt a hand of 3 cards from a standard deck. In how many ways can this be done?How many of the sets will contain at least 1 ace among the 3 cards?What is the probability of at least 1 ace among the 3 cards you are dealt?
Question
A standard deck of playing cards has 52 cards divided equally into four suits: hearts , diamonds , clubs , and spades . Within each suit, there are 13 cards: one ace, the numbers from 2 to 10, a jack, queen, and king.Suppose you are dealt a hand of 3 cards from a standard deck. In how many ways can this be done?How many of the sets will contain at least 1 ace among the 3 cards?What is the probability of at least 1 ace among the 3 cards you are dealt?
Solution
To answer the first question, we need to calculate the number of combinations of 3 cards from a 52-card deck. This can be done using the combination formula C(n, k) = n! / [k!(n-k)!], where n is the total number of items, k is the number of items to choose, and "!" denotes factorial.
So, C(52, 3) = 52! / [3!(52-3)!] = 22,100 ways.
For the second question, we need to calculate the number of ways to get at least 1 ace. This can be done by summing the number of ways to get exactly 1, 2, or 3 aces.
For exactly 1 ace: C(4, 1) * C(48, 2) = 4 * 1,128 = 4,512 ways. For exactly 2 aces: C(4, 2) * C(48, 1) = 6 * 48 = 288 ways. For exactly 3 aces: C(4, 3) = 4 ways.
So, the total number of ways to get at least 1 ace is 4,512 + 288 + 4 = 4,804 ways.
For the third question, the probability of getting at least 1 ace is the number of favorable outcomes (at least 1 ace) divided by the total number of outcomes (any 3 cards).
So, the probability is 4,804 / 22,100 = 0.217, or approximately 21.7%.
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