What is the probability

Question

Richard has just been given a 4-question multiple-choice quiz in his history class. Each question has four answers, of which only one is correct. Since Richard has not attended class recently, he doesn't know any of the answers. Assuming that Richard guesses on allfour questions, find the indicated probabilities. (Round your answers to three decimal places.)

(a) What is the probability that he will answer all questions correctly?
 

(b) What is the probability that he will answer all questions incorrectly?
 

(c) What is the probability that he will answer at least one of the questions correctly? Compute this probability two ways. First, use the rule for mutually exclusive events and the probabilities shown in the binomial probability distribution table.
 

Then use the fact that P(r ≥ 1) = 1 − P(r = 0).
 

Compare the two results. Should they be equal? Are they equal? If not, how do you account for the difference?

They should be equal, but may not be due to table error.They should not be equal, but are equal.    They should be equal, but differ substantially.They should be equal, but may differ slightly due to rounding error.

(d) What is the probability that Richard will answer at least half the questions correctly?

 

Answer 1

(a) The probability of answering each question correctly is p=0.25, so to answer all 4 questions correctly the probability is p⁴=1/256=0.00390625 (about 0.4%).

(b) The probability of incorrectly answering each question is p=0.75. Probability of total failure=81/256=0.31640625 (about 32%).

(c) (b) measured total failure which is P(r=0), so P(r≥1)=1-P(r=0)=1-81/256=175/256=0.68359375 (about 68%).

If p is the probability of an event occurring n times then:

(p+1-p)ⁿ=1 represents all possible outcomes, which must of course be 1 (100%). Let q=1-p.

If this expanded using the binomial expansion and n=4, we get:

(p+q)⁴=p⁴+4p³q+6p²q²+4pq³+q⁴=1.

Each term in the series represents a particular outcome. p⁴ represents 4 successes; 4p³q represents 3 successes and 1 failure; 6p²q² represents 2 successes and 2 failures; 4pq³ represents 1 success and 3 failures; and q⁴ represents total failure. So we would expect at least 1 success in all but the last term. If we call the result of adding these 4 outcomes R, then R=1-q⁴.

If we plug p=0.25 (q=0.75) into the expansion we get:

1/256+3/64+27/128+27/64=(1+12+54+108)/256=175/256, which is the same as 1-P(r=0). However, if we use tables we could get rounding errors depending on the accuracy of the tables.

(d) Half the questions is two questions, so we want the probabilities of 2 or more, P(r≥2).

Since the expansion of the terms has already been calculated, we can simply add the relevant terms:

P(r=2)+P(r=3)+P(r=4)=p⁴+4p³q+6p²q²=

1-(4pq³+q⁴)=1-(27/64+81/256)=

1-(108+81)/256=67/256=0.26171875 (about 26%). We would get the same result if we added the first three terms.