Mathematics: 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

its on googleStep-by-step explanation:dont ask kinder gardeners Which graph best represents the line y=-1/5x+2? ok...

An alcohol awareness task force at a Big-Ten university

shelovejaylocs

20.01.2022

(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.

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)

$=\frac{30}{100}\\=0.30$

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

$=\frac{110+100-30}{200}\\=\frac{180}{200}\\=0.90$

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

$=\frac{30}{200}\\=0.15$

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

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