Respuesta :
Answer:
0.09 = 9% probability that the student is proficient in neither reading nor mathematics
Step-by-step explanation:
We solve this question treating the events as Venn probabilities.
I am going to say that:
Event A: A student is proficient in reading.
Event B: A student is proficient in mathematics.
A total of 82% of the students were found to be proficient in reading
This means that [tex]P(A) = 0.82[/tex]
74% were found to be proficient in mathematics
This means that [tex]P(B) = 0.74[/tex]
65% were found to be proficient in both reading and mathematics.
This means that [tex]P(A \cap B) = 0.65[/tex]
What is the probability that the student is proficient in neither reading nor mathematics?
This is:
[tex]P = 1 - P(A \cup B)[/tex]
In which
[tex]P(A \cup B) = P(A) + P(B) - P(A \cap B)[/tex]
With the values that we have:
[tex]P(A \cup B) = 0.82 + 0.74 - 0.65 = 0.91[/tex]
Then
[tex]P = 1 - P(A \cup B) = 1 - 0.91 = 0.09[/tex]
0.09 = 9% probability that the student is proficient in neither reading nor mathematics
