Respuesta :
Answer:
There is a 2% probability that the student is proficient in neither reading nor mathematics.
Step-by-step explanation:
We solve this problem building the Venn's diagram of these probabilities.
I am going to say that:
A is the probability that a student is proficient in reading
B is the probability that a student is proficient in mathematics.
C is the probability that a student is proficient in neither reading nor mathematics.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a student is proficient in reading but not mathematics and [tex]A \cap B[/tex] is the probability that a student is proficient in both reading and mathematics.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So
[tex](A \cup B) + C = 1[/tex]
In which
[tex](A \cup B) = a + b + (A \cap B)[/tex]
65% were found to be proficient in both reading and mathematics.
This means that [tex]A \cap B = 0.65[/tex]
78% were found to be proficient in mathematics
This means that [tex]B = 0.78[/tex]
[tex]B = b + (A \cap B)[/tex]
[tex]0.78 = b + 0.65[/tex]
[tex]b = 0.13[/tex]
85% of the students were found to be proficient in reading
This means that [tex]A = 0.85[/tex]
[tex]A = a + (A \cap B)[/tex]
[tex]0.85 = a + 0.65[/tex]
[tex]a = 0.20[/tex]
Proficient in at least one:
[tex](A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98[/tex]
What is the probability that the student is proficient in neither reading nor mathematics?
[tex](A \cup B) + C = 1[/tex]
[tex]C = 1 - (A \cup B) = 1 - 0.98 = 0.02[/tex]
There is a 2% probability that the student is proficient in neither reading nor mathematics.
