Question

word problem involving the probability of a union or an intersection

eric has a 44% chance of receiving an a grade in chemistry, a 36% chance of receiving an a grade in mathematics, and a 46% chance of receiving an a grade in chemistry or mathematics (or both). find the probability that he receives a grades in both chemistry and mathematics.

write your answer as a decimal (not as a percentage). (if necessary, consult a list of formulas.)

Explanation:

Define the given probabilities

We represent the events and their given probabilities. Let:

• \(C\) be the event that Eric receives an A grade in chemistry.
• \(M\) be the event that Eric receives an A grade in mathematics.

Using the Probability of Events concept, we convert the percentages to decimals:

• \(P(C) = 44\% = 0.44\)
• \(P(M) = 36\% = 0.36\)
• \(P(C \cup M) = 46\% = 0.46\) (the probability of receiving an A in chemistry or mathematics, or both)

Apply the addition rule

Using the Addition Rule of Probability, we relate the union and intersection of two events:

$$P(C \cup M) = P(C) + P(M) - P(C \cap M)$$

Solve for the intersection

We rearrange the formula to solve for the target probability, \(P(C \cap M)\):

$$P(C \cap M) = P(C) + P(M) - P(C \cup M)$$

Substitute the known values:

$$P(C \cap M) = 0.44 + 0.36 - 0.46$$

$$P(C \cap M) = 0.80 - 0.46 = 0.34$$

Answer:

Eric has a 44% chance of receiving an A grade in chemistry, a 36% chance of receiving an A grade in mathematics, and a 46% chance of receiving an A grade in chemistry or mathematics (or both). Find the probability that he receives A grades in both chemistry and mathematics.

Write your answer as a decimal (not as a percentage).

<blank>0.34</blank>