Respuesta :
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 [tex]P(A) = 0.48, P(A \cap B) = 0.25[/tex]
Assume that 50% of the students are male.
This means that [tex]P(B) = 0.5[/tex]
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:
[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]
Considering the values we were given:
[tex]P(A \cup B) = 0.48 + 0.5 - 0.25 = 0.73[/tex]
0.73 = 73% probability that a randomly selected student is either male or would admit to lying to a teacher, during the past year.
