Question

learning goal from lesson 20.1
i can construct and interpret two - way frequency tables of data
when two categories are associated with each object. i can
starting...
determine independence of events using a two - way table as a
sample space. i can approximate conditional probabilities using a
two - way table as a sample space.
lesson 20.1 checkpoint
■ once you have completed the above problems and checked your solutions, complete the lesson checkpoint
below.
■ complete the lesson reflection above by circling your current understanding of the learning goal.
a researcher collected data from 120 students. the two - way frequency table shows the number of
students who passed and failed an exam and the number of students who got more or less than 6 hours
of sleep the night before.

1. what is the probability that a student who got more than 6 hours of sleep passed the exam?

■ a. $\frac{3}{4}$
■ b. $\frac{41}{50}$
■ c. $\frac{17}{20}$
■ d. $\frac{75}{82}$

1. you randomly draw a card from a standard deck of playing cards. let a be the event that the card is

an ace, let b be the event that the card is black, and let c be the event that the card is a club. find the
specified probability as a fraction.
a) $p(a|b)$
b) $p(b|c)$
■ a. $\frac{1}{2}$
■ a. 1
■ b. $\frac{1}{13}$
■ b. $\frac{1}{4}$
■ c. $\frac{1}{26}$
■ c. $\frac{1}{26}$

Explanation:

1. For the first question:

Step1: Recall the formula for conditional probability

The formula for conditional probability \(P(X|Y)=\frac{n(X\cap Y)}{n(Y)}\). Here, \(X\) is the event of passing the exam and \(Y\) is the event of getting more than 6 hours of sleep.

Step2: Identify the values from the table

From the table, \(n(X\cap Y) = 75\) (number of students who got more than 6 hours of sleep and passed the exam) and \(n(Y)=82\) (number of students who got more than 6 hours of sleep).
So, \(P(\text{passed}|\text{more than 6 hours of sleep})=\frac{75}{82}\)

2. For part (a) of the second question:

Step1: Recall the formula for conditional probability

\(P(A|B)=\frac{P(A\cap B)}{P(B)}\). In a standard deck of \(n = 52\) cards, \(P(B)=\frac{26}{52}\) (26 black cards) and \(P(A\cap B)=\frac{2}{52}\) (2 black aces: ace of spades and ace of clubs)

Step2: Calculate the probability

\(P(A|B)=\frac{\frac{2}{52}}{\frac{26}{52}}=\frac{2}{26}=\frac{1}{13}\)

2. For part (b) of the second question:

Step1: Recall the formula for conditional probability

\(P(B|C)=\frac{P(B\cap C)}{P(C)}\). Since all clubs (\(n(C) = 13\)) are black (\(B\cap C=C\)), \(P(B\cap C)=\frac{13}{52}\) and \(P(C)=\frac{13}{52}\)

Step2: Calculate the probability

\(P(B|C)=\frac{\frac{13}{52}}{\frac{13}{52}} = 1\)

Answer:

1. D. \(\frac{75}{82}\)
2. a) B. \(\frac{1}{13}\)

b) A. \(1\)