Out of the students in a class, 50% are geniuses, 60% love chocolate, and 30% fall into both categories. Determine the probability that a randomly selected student from the class is neither a genius nor a chocolate lover.

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:

In which a is the probability that a student is a genius but does not like chocolate and is the probability that a student is both a genius and likes chocolate.

By the same logic, we have that:

30% fall into both categories.

This means that

60% love chocolate

This means that . So

50% are geniuses

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

In which

We want C, so

20% probability that a randomly selected student from the class is neither a genius nor a chocolate lover.

answer: 88

step-by-step explanation:

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p