Statistics Question?? Help Me Please!

Statistics Question?? Help Me Please!



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 all four 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).





Posted Answers:

n=4 (number of questions) 
p = 1/4 (probability of a correct answer for a single question since there are 4 choices) 
r= number of correct answers out of 4 questions 

This is binomial probability 

a) 
P(r=4) = 4C4 (1/4)^4 (3/4)^(4-4) 
= (1/4)^4 
= 1/256 


b) 
P(r=0) = 4C0 (1/4)^0 (3/4)^4 
= (3/4)^4 
= 81/256 


c) 
Method 1: 
P( r ≥ 1) = P(r=1)+P(r=2)+P(r=3)+P(r=4) 

P(r=1) = 4C1 (1/4)^1 (3/4)^3 = 108/256 
P(r=2) = 4C2 (1/4)^2 (3/4)^2 = 54 /256 
P(r=3) = 4C3 (1/4)^3 (3/4)^1 = 12/256 
P(r=4) = 4C4 (1/4)^4 (3/4)^0 = 1/256 

add: 
P(at least one correct) = (108+54+12+1)/256 = 175/256 = 0.6836 

Method 2: 
P( r ≥ 1) = 1- P(r=0) 
P(r=0) = 81/256 (question 2) 

1-P(r=0) = 1-81/256 =175/256


 n=4 (number of questions) 
p = 1/4 (probability of a correct answer for a single question since there are 4 choices) 
r= number of correct answers out of 4 questions 

This is binomial probability 

a) 
P(r=4) = 4C4 (1/4)^4 (3/4)^(4-4) 
= (1/4)^4 
= 1/256 


b) 
P(r=0) = 4C0 (1/4)^0 (3/4)^4 
= (3/4)^4 
= 81/256 


c) 
Method 1: 
P( r ≥ 1) = P(r=1)+P(r=2)+P(r=3)+P(r=4) 

P(r=1) = 4C1 (1/4)^1 (3/4)^3 = 108/256 
P(r=2) = 4C2 (1/4)^2 (3/4)^2 = 54 /256 
P(r=3) = 4C3 (1/4)^3 (3/4)^1 = 12/256 
P(r=4) = 4C4 (1/4)^4 (3/4)^0 = 1/256 

add: 
P(at least one correct) = (108+54+12+1)/256 = 175/256 = 0.6836 

Method 2: 
P( r ≥ 1) = 1- P(r=0) 
P(r=0) = 81/256 (question 2) 

1-P(r=0) = 1-81/256 =175/256