An alcohol awareness task force at a Big-Ten university sampled 200 students after the midterm to ask them whether they went bar hopping the weekend before the midterm or spent the weekend studying, and whether they did well or poorly on the midterm. The following result was obtained. Did Well on Midterm Did Poorly on MidtermStudying for Exam 80 20Went Bar Hopping 30 70a. What is the probability that a randomly selected student who went bar hopping did well on the midterm?b. What is the probability that a randomly selected student did well on the midterm or went bar hopping the weekend before the midterm?c. What is the probability that a randomly selected student did well on the midterm and also went bar hopping the weekend before the midterm?

(a) The probability that a student who went bar hopping did well on the midterm is 0.30.

(b) The probability that a student did well on the midterm or went bar hopping is 0.90.

(c) The probability that a student did well on the midterm and also went bar hopping is 0.15.

Step-by-step explanation:

The data provided is:

                          Did well (W)   Did Poorly (P)   TOTAL (T)

Studying (S)              80                20                  100

Bar-hopping (B)        30                70                  100

TOTAL (T)                 110                90                 200

(a)

Compute the probability that a randomly selected student who went bar hopping did well on the midterm as follows:

P (W|B) = n (W ∩ B) ÷ n (B)

Thus, the probability that a student who went bar hopping did well on the midterm is 0.30.

(b)

Compute the probability that a randomly selected student did well on the midterm or went bar hopping the weekend before the midterm as follows:

P (W ∪ B) = [n (W) + n (B) - n (W ∩ B)] ÷ N

Thus, the probability that a student did well on the midterm or went bar hopping is 0.90.

(c)

Compute the probability that a randomly selected student did well on the midterm and also went bar hopping the weekend before the midterm as follows:

P (W ∩ B) = n (W ∩ B) ÷ N

Thus, the probability that a student did well on the midterm and also went bar hopping is 0.15.

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

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step-by-step explanation: