Question

chapter 3: 3.3 two basic rules of probability homework
score: 11/14 answered: 6/8
question 7
in a large school, it was found that 62% of students are taking a math class, 80% of student are taking an english class, and 55% of students are taking both.
find the probability that a randomly selected student is taking a math class or an english class. write your answer as a decimal, and round to 2 decimal places if necessary.
find the probability that a randomly selected student is taking neither a math class nor an english class. write your answer as a decimal, and round to 2 decimal places if necessary.

Explanation:

Step1: Recall the formula for the probability of the union

The formula for $P(A\cup B)$ is $P(A)+P(B)-P(A\cap B)$. Let $A$ be the event that a student is taking a math - class and $B$ be the event that a student is taking an English class. Given $P(A) = 0.62$, $P(B)=0.80$, and $P(A\cap B)=0.55$.
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Step2: Calculate $P(A\cup B)$

Substitute the given values into the formula:
$P(A\cup B)=0.62 + 0.80-0.55$
$P(A\cup B)=0.87$

Step3: Recall the complement rule

The probability of the complement of an event $E$, denoted as $P(\overline{E})$, is given by $P(\overline{E})=1 - P(E)$. Here, $E$ is the event that a student is taking a math class or an English class.
Let $E = A\cup B$. Then the probability that a student is taking neither a math class nor an English class is $P(\overline{A\cup B})$.
$P(\overline{A\cup B})=1 - P(A\cup B)$

Step4: Calculate $P(\overline{A\cup B})$

Since $P(A\cup B)=0.87$, then $P(\overline{A\cup B})=1 - 0.87=0.13$

Answer:

The probability that a randomly - selected student is taking a math class or an English class is $0.87$.
The probability that a randomly - selected student is taking neither a math class nor an English class is $0.13$.