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A pair of aces. What is the probability of drawing two aces in two random draws without replacement from a full deck of cards?

Short Answer

Expert verified
The probability is \(\frac{1}{221}\).

Step by step solution

01

- Understand the Problem

Determine the total number of aces and the total number of cards in the deck. A deck has 52 cards and 4 aces.
02

- Calculate the Probability of Drawing the First Ace

The probability of drawing an ace on the first draw is \(\frac{4}{52}\).
03

- Calculate the Probability of Drawing the Second Ace

After drawing the first ace, there are now 51 cards left and 3 aces remaining. The probability of drawing a second ace is then \(\frac{3}{51}\).
04

- Multiply the Probabilities

Multiply the probabilities of both events to get the overall probability of drawing two aces in succession. This is \(\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

card games probability
In probability theory, card games are a fantastic way to explore and understand different probability concepts. Let's delve into how probability works with a traditional deck of cards. A standard deck contains 52 cards, divided into 4 suits (hearts, diamonds, clubs, and spades), each containing 13 cards. Probabilities in card games are usually calculated by considering the ratio of favorable outcomes to the total number of possible outcomes.

In the case of drawing two aces without replacement, we start by recognizing that there are initially 4 aces in the 52-card deck. After the first ace is drawn, there are only 51 cards left, with 3 remaining aces. By calculating the sequential probabilities and multiplying them, you can find the overall chance of a specific sequence occurring during the game.
combinatorics
Combinatorics is a branch of mathematics dealing with counting, combination, and permutation of sets. Understanding basic combinatorics can significantly simplify finding probabilities, especially in scenarios like drawing cards from a deck.

When calculating the probability of drawing two aces in two tries, combinatorics help us determine how many ways we can select the cards. For example, before any card is drawn, there are 52 cards with 4 aces. Once a card is selected, we remove it from our calculations, considering fewer cards and aces remain. This understanding allows us to construct accurate mathematical models for diverse card-drawing problems.
conditional probability
Conditional probability is the probability of an event occurring given that another event has already occurred. In card games, this concept is crucial, especially for events that take place in succession without replacement.

In the problem of drawing two aces consecutively, the probability of the second draw depends on the first draw. Initially, the probability of drawing an ace is \(\frac{4}{52}\). However, after drawing one ace, the conditions change: there are now 51 remaining cards and only 3 aces left. Hence, the probability of the second draw being an ace is \(\frac{3}{51}\).

By multiplying these conditional probabilities, we can calculate the overall probability of drawing two aces in succession: \(\frac{4}{52} \times \frac{3}{51} = \frac{1}{221}\).
deck of cards
A standard deck of cards is a foundational element in probability exercises and card games. It consists of 52 cards divided into 4 suits: hearts, diamonds, clubs, and spades. Each suit contains 13 cards, numbered from 2 to 10, and includes a jack (J), queen (Q), king (K), and ace (A). Understanding the structure of a deck is essential, as it lays the groundwork for calculating probabilities.

Different problems, like drawing specific cards (such as aces) or combinations of cards, rely heavily on knowing the composition of the deck. For instance, knowing there are exactly four aces in a deck helps us accurately calculate probabilities involving aces. This understanding is crucial for more complex probability exercises and real-world applications.

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Most popular questions from this chapter

Predicting the rate of mutation based on the Poisson probability distribution function. The evolutionary process of amino acid substitution in proteins is sometimes described by the Poisson probability distribution function. The probability \(p_{s}(t)\) that exactly \(s\) substitutions at a given amino acid position occur over an evolutionary time \(t\) is $$ p_{s}(t)=\frac{e^{-\lambda t}(\lambda t)^{s}}{s !}, $$ where \(\lambda\) is the rate of amino acid substitution per site per unit time. Fibrinopeptides evolve rapidly: \(\lambda_{F}=9.0\) substitutions per site per \(10^{9}\) years. Lysozyme is intermediate: \(\lambda_{L} \approx 1.0\). Histones evolve slowly: \(\lambda_{H}=0.010\) substitutions per site per \(10^{9}\) years. (a) What is the probability that a fibrinopeptide has no mutations at a given site in \(t=1\) billion years? (b) What is the probability that lysozyme has three mutations per site in 100 million years? (c) We want to determine the expected number of mutations \(\langle s\rangle\) that will occur in time \(t\). We will do this in two steps. First, using the fact that probabilities must sum to one, write \(\alpha=\sum_{s=0}^{\infty}(\lambda t)^{s} / s !\) in a simpler form. (d) Now write an expression for \(\langle s\rangle\). Note that $$ \sum_{s=0}^{\infty} \frac{s(\lambda t)^{s}}{s !}=(\lambda t) \sum_{s=1}^{\infty} \frac{(\lambda t)^{s-1}}{(s-1) !}=\lambda t \alpha $$ (e) Using your answer to part (d), determine the ratio of the expected number of mutations in a fibrinopeptide to the expected number of mutations in histone protein, \(\langle s\rangle_{\mathrm{fib}} /\langle s\rangle_{\mathrm{his}}[6]\).

Probabilities of sequences. Assume that the four bases A, C, T, and G occur with equal likelihood in a DNA sequence of nine monomers. (a) What is the probability of finding the sequence AAATCGAGT through random chance? (b) What is the probability of finding the sequence AAAAAAAAA through random chance? (c) What is the probability of finding any sequence that has four A's, two T's, two G's, and one C, such as that in (a)?

Combining independent probabilities. You have a fair six-sided die. You want to roll it enough times to ensure that a 2 occurs at least once. What number of rolls \(k\) is required to ensure that the probability is at least \(2 / 3\) that at least one 2 will appear?

Predicting compositions of independent events. Suppose you roll a fair six- sided die three times. (a) What is the probability of getting a 5 twice from all three rolls of the dice? (b) What is the probability of getting a total of at least two 5 's from all three rolls of the die?

DNA synthesis. Suppose that upon synthesizing a molecule of DNA, you introduce a wrong base pair, on average, every 1000 base pairs. Suppose you synthesize a DNA molecule that is 1000 bases long. (a) Calculate and draw a bar graph indicating the yield (probability) of each product DNA, containing 0,1 , 2 , and 3 mutations (wrong base pairs). (b) Calculate how many combinations of DNA sequences of 1000 base pairs contain exactly 2 mutant base pairs. (c) What is the probability of having specifically the 500 th base pair and the 888 th base pair mutated in the pool of DNA that has only two mutations? (d) What is the probability of having two mutations side-by-side in the pool of DNA that has only two mutations?

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