she needs at least 4 out of the remaining 8 questions correct. So, that has probability
This topic is discussed here:
http://math.stackexchange.com/questions/853407/4-heads-in-8-tosses
a multiple choice quiz has 10 questions. each question has five possible answers.carman is certain that she knows the correct answers for questions 1 and 4. if she guesses the other questions, determine the probability that she gets 60 % on the quiz ?
i know this question involves binomial distribution but im not sure with the correct format and order that im supposed to put the question in ?
2 answers
The probability of any one getting correct with five possible answers is .2, and to make a 60 she has to get four more right.
P(4 correct)=8C4*(.2)^4(.8)^4=
= 8*7*6*5/4*3*2*1 (.2)^4 * (.8)^4
= 7*2*5 * (.2)^4 (.8)^4
= 70 * 0.00065536
= 0.0458752
Now if you wanted to calculate the probability of getting at least a 60 , then you have to add the Pr of getting 70, 80, 90, 100
Pr(70:5 more right)=8C5 (.2)^5(.8)^3
Pr(80:6 right)=8C6 (.2)^6 (.8)^2
Pr(90:7 right)=8C7 (.2)^7 (.8)
Pr (100: 8 more right)= .2^8
P(4 correct)=8C4*(.2)^4(.8)^4=
= 8*7*6*5/4*3*2*1 (.2)^4 * (.8)^4
= 7*2*5 * (.2)^4 (.8)^4
= 70 * 0.00065536
= 0.0458752
Now if you wanted to calculate the probability of getting at least a 60 , then you have to add the Pr of getting 70, 80, 90, 100
Pr(70:5 more right)=8C5 (.2)^5(.8)^3
Pr(80:6 right)=8C6 (.2)^6 (.8)^2
Pr(90:7 right)=8C7 (.2)^7 (.8)
Pr (100: 8 more right)= .2^8
1 answer