Question

2.1: The probability that a visit to a local restaurant results in neither buying an appetizer nor a dessert is 55%. Of those coming to the restaurant, 25% buy an appetizer and 30% buy dessert. What is the probability that a visit leads to buying both an appetizer and dessert?

2.2: In a high school class, 35% of the students take Spanish as a foreign language, 15% take French as a foreign language, and 40% take at least one of these languages. What is the probability that a randomly chosen student takes French given that the student takes Spanish?

Answer #1

2.1)

Let A = buy an appetizer

B = Buy dessert.

P(A) = 0.25

P(B) = 0.30

P(A OR B)' = Probability that neither buying an appetizer nor dessert = 0.55

Therefore,

P(A OR B) = 1 - P(A OR B)'

= 1 - 0.55

P( A OR B) = 0.45

P( A AND B) = ?

Using addition rule,

P(A OR B) = P(A) + P(B) - P( A AND B)

0.45 = 0.25 + 0.30 - P(A AND B)

P(A AND B) = 0.25 + 0.30 - 0.45

= 0.1

**The probability of buying both an appetizer and dessert
= 0.10**

2.2)

Let A = Spanish as foreign language

B = French as foreign language.

P(A) = 0.35

P(B) = 0.15

P(A OR B) = 0.40

P(B | A) = ?

Using addition rule,

P(A OR B) = P(A) + P(B) - P( A AND B)

0.40 = 0.35 + 0.15 - P(A AND B)

P(A AND B) = 0.35 + 0.15 - 0.40

P(A AND B) = 0.10

Therefore,

P(B | A) = P(A AND B) / P(A)

= 0.10 / 0.35

= 0.2857

**Probability of student takes French given that the
student takes Spanish = 0.2857**

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