Question

Consider a multiple choice exam with 10 questions, and with 4 possible answers to each question (labeled A, B, C, and D). 1a. What is the probability of getting a perfect score just by guessing? 1b. What is the probability of getting at least one question correct, just by guessing? Explain your thought process briefly. 2. Each RUID is 9-digits long, but the 4th and 5th digits must both be 0. How many different RUIDs can there be?

Answer 1

Answer:

(1a) 9.54e-07 (1b) 0.9437 (2) $10^8$

Step-by-step explanation:

We have n = 10 questions. For each question in the multiple choice exam, we can choose the correct answer with probability p = 0.25 or an incorrect answer with probability q = 0.75 (just by guessing). Besides is reasonable to believe that questions are independent. Let X be the random variable that represents the number of correct answers (just by guessing) observed during the n = 10 questions, so, X have a binomial distribution.

(a) The probability of getting a perfect score just by guessing is

$P(X = 10) = 10C10(0.25)^{10}(0.75)^{10-10} = 9.54e-07$

(b) The probability of getting at least one question correct, just by guessing

$P(X\geq1) = \sum_{x=1}^{x=10}10Cx(0.25)^{x}(0.75)^{(10-x)} = 0.9437$

2. Each RUID is 9-digits long, there are 10 digits, each of the nine positions for a digit of the RUID has 10 posibilities except the 4th and 5th positions which must be both 0. For the multiplication rule there can be $(10)^3(1)(1)(10)^5 = 10^8$ different RUIDS.