(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.