Respuesta :
Answer:
(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) Â Â Â Â Â Â Â 80 Â Â Â Â Â Â Â Â 20 Â Â Â Â Â Â Â Â Â 100
Bar-hopping (B) Â Â Â Â 30 Â Â Â Â Â Â Â Â 70 Â Â Â Â Â Â Â Â Â 100
TOTAL (T) Â Â Â Â Â Â Â Â 110 Â Â Â Â Â Â Â Â 90 Â Â Â Â Â Â Â Â 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)
      [tex]=\frac{30}{100}\\=0.30[/tex]
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
        [tex]=\frac{110+100-30}{200}\\=\frac{180}{200}\\=0.90[/tex]
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
        [tex]=\frac{30}{200}\\=0.15[/tex]
Thus, the probability that a student did well on the midterm and also went bar hopping is 0.15.
