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Chapter 16: Problem 18

Two cards are drawn at random from a shuffled deck. (a) What is the probability that at least one is a heart? (b) If you know that at least one is a heart, what is the probability that both are hearts?

Short Answer

Expert verified
(a) \( \frac{195}{442} \(b) \( \frac{2}{15} \)

Step by step solution

01

- Total number of outcomes

Calculate the total number of ways to draw 2 cards from a deck of 52. This is given by the combination formula \(\binom{52}{2}\). Use the formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), where \(n = 52\) and \(r = 2\). \[ \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 1326 \]
02

- Successful outcomes for 'at least one heart'

Calculate the number of ways to draw at least one heart. Instead of directly calculating the number of successful outcomes, use the complementary approach: calculate the number of ways to draw 0 hearts, and subtract from the total. There are \( \binom{39}{2} \) ways to draw 2 non-heart cards (39 non-heart cards in a deck). \[ \binom{39}{2} = \frac{39 \times 38}{2 \times 1} = 741 \] The number of ways to draw at least one heart is then \[ 1326 - 741 = 585 \]
03

- Probability of 'at least one heart'

The probability that at least one card is a heart is the ratio of successful outcomes to the total number of outcomes. \[ P(\text{at least one heart}) = \frac{585}{1326} \] After simplifying, we get: \[ P(\text{at least one heart}) = \frac{195}{442} \]
04

- Successful outcomes for 'both hearts'

Calculate the number of ways to draw 2 hearts from the 13 hearts available. Use the combination formula \( \binom{13}{2}\). \[ \binom{13}{2} = \frac{13 \times 12}{2 \times 1} = 78 \]
05

- Conditional probability

The conditional probability that both cards are hearts given that at least one is a heart is found using the conditional probability formula: \(\frac{P(\text{both hearts})}{P(\text{at least one heart})}\). Substitute the values from the previous steps: \[ P(\text{both hearts | at least one heart}) = \frac{\frac{78}{1326}}{\frac{585}{1326}} = \frac{78}{585} = \frac{26}{195} = \frac{2}{15} \]

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Key Concepts

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

Combinatorics
Combinatorics is a field of mathematics that studies the counting, arrangement, and combination of objects. One of the common tools used in combinatorics is the combination formula, denoted as \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]. It helps us calculate the number of ways to choose a subset of \( r \) elements from a larger set of \( n \) elements without considering the order. For example, to find the total number of ways to draw 2 cards from a deck of 52, we use \[ \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 1326 \] combinations.
Conditional Probability
Conditional probability measures the likelihood of an event occurring given that another event has already occurred. The formula for conditional probability is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]. In our exercise, after finding the probability that at least one card is a heart, we determine the conditional probability that both cards are hearts given this condition. Using the given numbers: \[ P(\text{both hearts | at least one heart}) = \frac{\frac{78}{1326}}{\frac{585}{1326}} = \frac{78}{585} = \frac{2}{15} \]. This tells us how likely it is to draw two hearts if we already know one of the cards is a heart.
Complement Rule in Probability
The complement rule in probability states that the probability of an event occurring is 1 minus the probability of it not occurring: \[ P(A) = 1 - P(A^c) \]. This rule is useful for calculating probabilities of events indirectly. For instance, to find the probability of drawing at least one heart from two cards, we first find the probability of drawing no hearts and then subtract from 1. Calculating zero hearts from 39 non-heart cards: \[ \binom{39}{2} = \frac{39 \times 38}{2 \times 1} = 741 \]. Then, \[ P(\text{at least one heart}) = 1 - \frac{741}{1326} = \frac{195}{442} \].
Drawing Cards Probabilities
When calculating probabilities in card games, it’s important to account for all possible outcomes and consider combinations. For example, to find the probability of drawing 2 hearts out of a 52-card deck, we use combinations: \[ \binom{13}{2} = \frac{13 \times 12}{2 \times 1} = 78 \]. To then get the conditional probability of both being hearts when we know at least one is a heart, we use this result along with the complement rule-derived figures: \[ P(\text{both hearts | at least one heart}) = \frac{\frac{78}{1326}}{\frac{585}{1326}} = \frac{78}{585} = \frac{2}{15} \]. Understanding these steps ensures accurate calculations in card games and other probabilities.

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Most popular questions from this chapter

A card is selected from a shuffled deck. What is the probability that it is either a king or a club? That it is both a king and a club?

Show that adding a constant \(K\) to a random variable increases the average by \(K\) but does not change the variance. Show that multiplying a random variable by \(K\) multiplics both the average and the standard deviation by \(K\).

A student claims in Problem \(1.5\) that if one child is a girl, the probability that both are girls is \(\frac{1}{2}\). Lise appropriate sample spaces to show what is wrong with the following argument: It doesn't matter whether the girl is the older child or the younger; in either case the probabitity is \(\frac{1}{2}\) that the other child is a girl.

Define the sample variance by \(s^{2}=(1 / n) \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} .\) Show that the expected value of \(s^{2}\) is \([(n-1) / n] \sigma^{2} .\) Hints: Write $$ \begin{aligned} \left(x_{i}-\bar{x}\right)^{2} &=\left[\left(x_{i}-\mu\right)-(\bar{x}-\mu)\right]^{2} \\ &=\left(x_{i}-\mu\right)^{2}-2\left(x_{i}-\mu\right)(\bar{x}-\mu)+(\bar{x}-\mu)^{2} \end{aligned} $$ Find the average value of the first term from the definition of \(\sigma^{2}\) and the average value of the third term from Problem 2, To find the average value of the middle term write $$ (\bar{x}-\mu)=\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}-\mu\right)=\frac{1}{n}\left[\left(x_{1}-\mu\right)+\left(x_{2}-\mu\right)+\cdots+\left(x_{n}-\mu\right)\right] $$ show by Problem \(7.12\) that $$ E\left[\left(x_{i}-\mu\right)\left(x_{j}-\mu\right)\right]=E\left(x_{i}-\mu\right) E\left(x_{j}-\mu\right)=0 \quad \text { for } \quad t \neq j $$ and evaluate \(E\left[\left(x_{i}-\mu\right)^{2}\right]\) (same as the first term). Collect terms to find $$ E\left(s^{2}\right)=\frac{n-1}{n} \sigma^{2} $$

Recall that two events \(A\) and \(B\) are called independent if \(p(A B)=p(A) \cdot p(B) .\) Similarly, two random variables \(x\) and \(y\) with probability functions \(f(x)\) and \(g(y)\) are called independent if the probability of \(x=x_{i}\) and \(y=y_{j}\) is \(f\left(x_{i}\right) \cdot g\left(y_{j}\right)\) for every pair of values of \(x\) and \(y\), that is, if the joint probability function for \(x, y\) is \(f(x) g(y)\). Show that if \(x\) anct \(y\) are independent, then the expectation or average of \(x y\) is \(E(x y)=E(x) \cdot E(y)=\mu_{x} \mu_{y}\).
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