Question 899049: a student is taking a multiple choice exam in which each question has 5 choices. assuming he has no knowledge of the correct answers to any of the questions, he has decided on a strategy in which he will place 5 balls marked ( A, B,C,D,E) into a box. he randomly selects one ball for each of the questions and replaces the ball back to the box. there are 4 questions.
a) what is the probability that he will obtain no marks in the exam?
b) what is the probability that he will get no more than 2 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) = .4096 0r 40.96%
2. P(x ≤ 2)= binomcdf(4, .2, 2) = .9728 0r 97.28%
TI syntax is binomcdf(n, p, largest x-value) for binomial ≤ cumulative probability
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