Question

Suppose that 48% of high school students would admit to lying at least once to a teacher during the past year and that 25% of students are male and would admit to lying at least once to a teacher during the past year.20 Assume that 50% of the students are male. What is the probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year? Be sure to show your work and indicate all the rules that you use to find your answer.

Answer 1

Answer:

0.73 = 73% probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year.

Step-by-step explanation:

I am going to treat these events as Venn probabilities, considering that:

Event A: Lying to the teacher.

Event B: Male

48% of high school students would admit to lying at least once to a teacher during the past year and that 25% of students are male and would admit to lying at least once to a teacher during the past year

This means that $P(A) = 0.48, P(A \cap B) = 0.25$

Assume that 50% of the students are male.

This means that $P(B) = 0.5$

What is the probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year?

This is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Considering the values we were given:

$P(A \cup B) = 0.48 + 0.5 - 0.25 = 0.73$

0.73 = 73% probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year.