Question

In a large school, it was found that 71% of students are taking a math class, 77% of student are taking an English class, and 58% 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.

Answer 1

Answer:

P (Math or English) = 0.90

Step-by-step explanation:

* 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 71%

- The probability of taking English class 77%

- The probability of taking both classes is 58%

∵ P(Math) = 71% = 0.71

∵ P(English) = 77% = 0.77

∵ P(Math and English) = 58% = 0.58

- 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) = 0.71 + 0.77 - 0.58 ⇒ simplify

∴ P(Math or English) =  0.90

* P(Math or English) = 0.90