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Guessing Answers Standard tests, such as the SAT, ACT, or Medical College Admission Test (MCAT), typically use multiple choice questions, each with five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to the first three questions. a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third is correct. That is, find P(WWC), where W denotes a wrong answer and C denotes a correct answer. b. Beginning with WWC, make a complete list of the different possible arrangements of two wrong answers and one correct answer, then find the probability for each entry in the list. c. Based on the preceding results, what is the probability of getting exactly one correct answer when the guesses are made?

Respuesta :

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!

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