Recall the formula for a number of combinations of n objects taken r at a time, where r ≤ n.
A
Practice makes perfect
In a standard deck of 52 cards there are 4 aces. We want to find the probability that a randomly dealt 5-card hand contains a pair of aces. To find it, we will use the theoretical probability.
P=Favorable Outcomes/Possible Outcomes
We start by finding the number of possible outcomes. This will be the number of combinations in which we can choose 5 out of 52 cards in a standard deck. The order in which we choose them is not important, since we consider them as a whole 5-card hand. Let's recall the formula for the number of combinations of n objects taken r at a time.
_nC_r = n!/(n-r)! * r!
There are 52 cards in a standard deck, so n= 52. Out of them, 5 cards are randomly selected. Therefore, we know that r = 5. Let's substitute these values and find the number of possible combinations.
Therefore, the number of possible outcomes is 2 598 960. Next, we will look for the number of favorable outcomes. We want to have a pair of aces in a dealt 5-card hand. Therefore, we need to look for the number of combinations of 2 aces taken out of 4, so that our 5-card hand contains a pair of aces.
We found there are 6 different combinations of a pair of aces that can be chosen. Notice that we want to have a 5-card hand. Let's find out how many more cards we need to draw.
5 - 2 = 3 cards remaining
We want to have exactly two aces in the dealt hand, therefore when choosing the 3 remaining cards we need to pick from cards other than aces.
52-4=48 cards to choose from
Therefore, we need to find the number of combinations of 3 cards chosen out of 48 cards other than aces. Therefore, let's evaluate _(48)C_3.
Notice that the choice of aces is independent from the choice of other cards. That enables us to use the Fundamental Counting Principle, since this principle is used to find the number of possible outcomes for a combination of independent events.
Fundamental Counting Principle
If an event A has n possible outcomes and an event B has m possible outcomes, then the total number of different outcomes for A and B combined is n * m.
Therefore, to find the total number of favorable outcomes, we need to multiply the number of combinations of aces and other cards.
Let's substitute the obtained values and evaluate the number of favorable outcomes.
6 * 17 296 = 103 776
We have enough information to calculate the desired probability.
Let's substitute the obtained values and evaluate the number of favorable outcomes.
6 * 17 296 = 103 776
We have enough information to calculate the desired probability.
The probability that a randomly dealt hand of 5 cards contains a pair of aces is about 4 %. This corresponds to answer A.