Question 898803: a student is taking a multiple choice exam in which each question has five choices. assuming that he has no knowledge of the correct answer to any of the questions, he has decided on a strategy in which he will place five balls (marked A,B,C,D,E)into a box. he randomly selects one ball for each question and replaces the ball in the box. the marking on the ball will determine his answer to the question. there are four multiple choice questions on the exam.
1. what is the probability that he will obtain no marks in the exam?
2. what is the probability that he will get no more than two questions correct?
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
p(correct) = 1/5 = .20, n = 4
TI syntax for P(x-value) is binompdf(n, p, x-value).
1.P(x = 0) = binompdf(4, .2, 0).
2. P(x ≤ 2)= binomcdf(4, .2, 2)
TI syntax is binomcdf(n, p, largest x-value) for binomial ≤ cumulative probability
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