Answer:
a) Probability of picking the first two answers wrong & the third answer correctly in that order, P(WWC) = 0.128
b) All possible outcomes = WWC, WCW, CWW
P(WWC) = 0.128; P(WCW) = 0.128; P(CWW) = 0.128
c) Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = 0.384
Step-by-step explanation:
P(Correct answer) = P(C) = number of correct options/total options available.
That is, P(C) = 1/5 = 0.2
P(Wrong answer) = P(W) = 1 - 0.2 = 0.8 or (number of wrong options)/(total options available) = 4/5 = 0.8
a) Using the multiple rule for independent events,
P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128
Probability of picking the first two answers wrong & the third answer correctly in that order = 0.128
b) All possible outcomes of picking two wrong answers and one right answer in whichever order for the first 3 questions are (WWC, WCW, CWW)
P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128
P(WCW) = P(W) × P(C) × P(W) = 0.8 × 0.2 × 0.8 = 0.128
P(CWW) = P(C) × P(W) × P(W) = 0.2 × 0.8 × 0.8 = 0.128
c) Using the addition rule for disjoint events,
Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = P(WWC) + P(WCW) + P(CWW) = 0.128 + 0.128 + 0.128 = 0.384
QED!
