Respuesta :
Answer:
20% probability that a randomly selected student from the class is neither a genius nor a chocolate lover.
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 a genius.
B is the probability that a men student likes chocolate.
C is the probability that a student is not a genius and does not like chocolate.
We have that:
[tex]A = a + (A \cap B)[/tex]
In which a is the probability that a student is a genius but does not like chocolate and [tex]A \cap B[/tex] is the probability that a student is both a genius and likes chocolate.
By the same logic, we have that:
[tex]B = b + (A \cap B)[/tex]
30% fall into both categories.
This means that [tex]A \cap B = 0.3[/tex]
60% love chocolate
This means that [tex]B = 0.6[/tex]. So
[tex]B = b + (A \cap B)[/tex]
[tex]0.6 = b + 0.3[/tex]
[tex]b = 0.3[/tex]
50% are geniuses
[tex]A = a + (A \cap B)[/tex]
[tex]0.5 = a + 0.3[/tex]
[tex]a = 0.2[/tex]
We know that either a student is at least a genius or likes chocolate, or a student is not a genius and does not like chocolate. 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) = 0.2 + 0.3 + 0.3 = 0.8[/tex]
We want C, so
[tex](A \cup B) + C = 1[/tex]
[tex]0.8 + C = 1[/tex]
[tex]C = 0.2[/tex]
20% probability that a randomly selected student from the class is neither a genius nor a chocolate lover.
