Question

in a large school, it was found that 68% of students are taking a math class, 76% of student are taking an english class, and 60% 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 principle of inclusion - exclusion for probability

The formula for \( P(M \cup E) \) (probability of taking math or English) is \( P(M \cup E)=P(M)+P(E)-P(M \cap E) \), where \( M \) is the event of taking a math class and \( E \) is the event of taking an English class.
Given \( P(M) = 0.68 \), \( P(E)=0.76 \), and \( P(M \cap E)=0.60 \).
Substitute the values into the formula: \( P(M \cup E)=0.68 + 0.76-0.60 \)
Calculate: \( 0.68+0.76 = 1.44 \), then \( 1.44 - 0.60=0.84 \).

Step2: Find the probability of neither event

The probability of neither \( M \) nor \( E \) is \( P(\overline{M \cup E})=1 - P(M \cup E) \)
We know \( P(M \cup E) = 0.84 \), so \( P(\overline{M \cup E})=1 - 0.84 = 0.16 \)

Answer:

For the probability of taking a math class or an English class: \( 0.84 \)
For the probability of taking neither a math class nor an English class: \( 0.16 \)