In a large school, it was found that 69% of students are taking a math class, 70% of student are taking an English class, and 50% of students are taking both. A. True
B. False

P(Math or English) = 0.89

Step-by-step explanation: This solution will only be applicable if finding the probability that a randomly selected student is taking a math class or an English class.

Lets study the meaning of or , and on probability. The use of the word or means that you are calculating the probability

that either event A or event B happened

Both events do not have to happen

The use of the word and, means that both event A and B have to happened

The addition rules are: # P(A or B) = P(A) + P(B) ⇒ mutually exclusive (events cannot happen

at the same time)

P(A or B) = P(A) + P(B) - P(A and B) ⇒ non-mutually exclusive (if they

have at least one outcome in common)

The union is written as A ∪ B or “A or B”.

The Both is written as A ∩ B or “A and B”

Lets solve the question

The probability of taking Math class 69%

The probability of taking English class 70%

The probability of taking both classes is 50%

P(Math) = 69% = 0.69

P(English) = 70% = 0.70

P(Math and English) = 50% = 0.50

To find P(Math or English) use the rule of non-mutually exclusive

P(A or B) = P(A) + P(B) - P(A and B)

P(Math or English) = P(Math) + P(English) - P(Math and English)

Lets substitute the values of P(Math) , P(English) , P(Math and English)

in the rule P(Math or English) = 0.69 + 0.70 - 0.50 = 0.89

P(Math or English) = 0.89

P(Math or English) = 0.89

This solution will only be applicable if we are to find the probability that a randomly selected student is taking a math class or an English class.

cheater

step-by-step explanation:

2 2/3 hours duhh silly

step-by-step explanation: