Question

Two cards are drawn from a shuffled deck. What is the probability that both are red? If ar least one is red, what is the probability that both are red? If one is a red ace, what is the probability that both are red?

Short answer

The probability that both cards are red is \( \frac{25}{102} \). Given at least one red card, the probability is \( \frac{25}{77} \). Given one is a red ace, the probability is \( \frac{25}{51} \).

Step-by-step solution

- Calculate the probability that both cards are red

First, calculate the probability of drawing one red card. There are 26 red cards (hearts and diamonds) in a standard 52-card deck. Probability = \ \ \[ \frac{26}{52} = \frac{1}{2} \] If the first card is red, then there are 25 red cards left out of the remaining 51 cards.Probability of second card being red = \ \ \[ \frac{25}{51} \] Thus, the probability that both cards are red is:\ \ \[ \frac{1}{2} \times \frac{25}{51} = \frac{25}{102} \]

- Calculate the probability that both are red given at least one is red

Use conditional probability. Let A be the event that both are red, and B be the event that at least one is red. We need to find P(A|B).Using the formula for conditional probability, \ \ \[ P(A|B) = \frac{P(A\cap B)}{P(B)} \] Since A is a subset of B, \ \ \[ P(A \cap B) = P(A) = \frac{25}{102} \] To find P(B), use complementary probability:Probability of no red cards (both black) = \ \ \[ \frac{26}{52} \times \frac{25}{51} = \frac{25}{102} \] Thus, \ \ \ \ \[ P(B) = 1 - \frac{25}{102} = \frac{77}{102} \] Then, \ \ \[ P(A|B) = \frac{\frac{25}{102}}{\frac{77}{102}} = \frac{25}{77} \]

- Calculate the probability that both are red given one is a red ace

There are 2 red aces in the deck. If one card is a red ace, there are 25 remaining red cards out of 51 cards since one red ace is already drawn. Thus, the probability that the other card is red = \ \ \[ \frac{25}{51} \]

Key concepts

conditional probability

Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. It's denoted as P(A|B), meaning the probability of event A occurring given that event B has occurred. To calculate it, you use the formula: \[ P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \] This concept helps in narrowing down the probability space by focusing only on specific known conditions.
For instance, in our exercise involving two cards: if you know at least one is red, the probability that both are red is simplified because you don't consider cases where both cards are black anymore. Calculating this involves understanding that event B (at least one red) changes the original sample space.

card probability

Card probability refers to the chances of drawing certain cards from a deck. A standard deck has 52 cards with 26 red cards (hearts, diamonds) and 26 black cards (spades, clubs).
In our problem, we first calculate the probability of drawing a red card. This is straightforward: \[ \frac{26}{52} = \frac{1}{2} \]
Once one red card is drawn, the deck has 51 remaining cards, 25 of which are red. So, the probability of drawing another red card: \[ \frac{25}{51} \]
This basic understanding of card distribution and subsequent changes in probabilities is key to solving more complex problems.

combinatorial probability

Combinatorial probability involves calculating probabilities where different combinations of events must be considered. It's useful when dealing with scenarios where order does not matter.
For example, in our exercise, after calculating the probability of both cards being red, we consider the combinations where at least one is red. This requires:

• First, finding the probability of at least one red card, which is easier by determining the complementary event (no red cards).
• Next, subtracting this probability from 1 to get the desired probability.

The key is to systematically consider all possible combinations and use the necessary formulas to calculate the probabilities of interest.

step-by-step solution

Here’s a breakdown of the step-by-step solution process for the given problem:

Step 1: Probability both cards are red:
Calculate the probability of first red card: \[ \frac{26}{52} = \frac{1}{2} \] Then, the second red card, given the first was red: \[ \frac{25}{51} \] The combined probability is: \[ \frac{1}{2} \times \frac{25}{51} = \frac{25}{102} \]

Step 2: Probability both are red given at least one is red:
Use conditional probability: \[ P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \] Since A is within B: \[ P(A \text{ and } B) = P(A) = \frac{25}{102} \] Calculate P(B) as: \[ 1 - P(\text{no red cards}) = 1 - \frac{25}{102} = \frac{77}{102} \] Then: \[ P(A|B) = \frac{\frac{25}{102}}{\frac{77}{102}} = \frac{25}{77} \]

Step 3: Probability both are red given one is a red ace:
With one red ace drawn: \[ \frac{25}{51} \]These steps simplify complex problems into manageable calculations.