Question

Two cards are drawn from a shuffled deck. What is the probability that both are aces? If you know that at least one is an ace, what is the probability that both are aces? If you know that one is the ace of spades, what is the probability that both are aces?

Short answer

The probabilities are: 1/221, 221/3309, and 1/17 respectively.

Step-by-step solution

Calculate the probability that both cards are aces

There are 4 aces in a standard 52-card deck. The probability of drawing the first ace is \(\frac{4}{52}\). After drawing one ace, there are 3 aces left out of 51 cards. Therefore, the probability of drawing a second ace is \(\frac{3}{51}\). Thus, the combined probability is the product of these two probabilities:\[\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}\]

Calculate the probability of both being aces given one is an ace

Given that at least one card is an ace, we first determine the total number of possible favorable outcomes. There are 4 aces among 52 cards, so the probability of picking one ace and another card from the remaining deck is \(\frac{4}{52} \text{ for the first ace and } \frac{48}{51}\) for the non-ace. We've already calculated the probability of two aces in Step 1, which is \(\frac{1}{221}\). Total ways to get at least one ace from both cards:\[1 - \text{probability that none of the cards is an ace} = 1 - \frac{48}{52} \times \frac{47}{51} = 1 - \frac{1888}{2652} = \frac{764}{2652} = \frac{191}{663}\]. Hence, the probability that both are aces given at least one is: \[\frac{\frac{1}{221}}{\frac{191}{663}} = \frac{663}{221 \times 191} = \frac{663}{4211} \text{ which simplifies to } \frac{221}{3309}\]

Calculate the probability of both being aces given one is the ace of spades

Given that the first card drawn is the ace of spades, we need to find the probability that the second card is also an ace. With the ace of spades already drawn, there are 3 aces left in a deck of 51 cards. This probability is\[\frac{3}{51} = \frac{1}{17}\]

Key concepts

conditional probability

Conditional probability is about finding the likelihood of an event given that another event has already happened. For example, in our exercise, we first calculate the probability of drawing two aces in a row. Given these two cards must be drawn from a deck of 52 cards, the probability for drawing the first ace is \(\frac{4}{52}\). After drawing one ace, notably, there are now 3 aces left and only 51 cards. Consequently, the probability of drawing a second ace is \(\frac{3}{51}\). Multiplying these together, we get \(\frac{1}{221}\).
Moving forward to conditional probabilities, the problem provides specific scenarios, such as 'given at least one card is an ace'. This is when conditional probability shines as it refines our sample space to figure out more restrictive conditions. Instead of computing from all outcomes, we now narrow it down. Therefore, we examine the probability of this particular scenario happening among the restricted outcomes. In practice, you may often see a formula for conditional probability:
\[P(A|B) = \frac{P(A \text{ and } B)}{P(B)}\]
Putting in simple terms, this means: the probability of event A happening given that event B has occurred, equals the probability that both events A and B happen divided by the probability of B alone happening.

combinatorics

Combinatorics is a branch of mathematics centered on counting, arrangements, and combinations of objects. When dealing with probabilities, we often use combinatorics to calculate the number of possible outcomes. Let’s break down our steps.
In our exercise, we ask about drawing aces from a deck of cards. First, we find the number of ways to draw aces: \(\frac{4}{52}\) for the first and \(\frac{3}{51}\) for the second. These fractions represent our favorable outcomes divided by the total possible outcomes, and the total probability is their product.
Next, the scenario changes to finding probabilities given known restrictions. If it's known that a drawn card is an ace, we must recalculate our total possible outcomes. Here, the idea called 'complementary counting' is applied: calculating the probability of something not happening and subtracting from 1. For instance, to get the probability that none of the cards are aces, we use:
\[1 - (\text{Probability that both cards are non-aces})\]
Breaking it down further, there are 48 non-aces in a deck, therefore the combined probability is \(\frac{48}{52} \times \frac{47}{51}\), which is then subtracted from 1 to refine our desired probability.

card probabilities

Card probabilities help in answering questions about drawing cards from a deck. For every problem involving a deck, notation is highly critical. First, each card drawn reduces the number of remaining cards, altering the total count impacting subsequent probabilities. This changing total must be kept in mind.
To illustrate, in our problem, we first calculate probabilities drawing without replacement. If the first card is an ace with probability \(\frac{4}{52}\), the second probability is now given one ace is removed: \(\frac{3}{51}\). Consequently: \(\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}\).
The problem involves another specific condition: knowing one of the cards is the Ace of Spades, which restricts our original possible choices. The probability of drawing another ace after this card is gone is now simplified to 3 remaining aces out of a 51-card deck: \(\frac{3}{51} = \frac{1}{17}\).
Recapping important tips:

• Always be clear about whether cards are drawn with or without replacement.
• Reducing total possible outcomes with every draw has a major impact.
• Account restrictions filtering the sample space when conditions are given.

Learning to appropriately apply these adjustments strengthens logical problem-solving and a meaningful grasp on card probabilities.