10 multiple choice questions each with three possible answers only one of which is correct. A student guesses the answer to every question.
(a)What is the probability that the student will answer the first six questions correctly and the rest incorrectly?
(b)What is the probability that the student will answer the 1st,2nd,4th,6th,7th,and 10th questions correctly and the rest incorrectly?
(c)How many ways can the student get exactly 6 correct answers?
(d)What is the probability that the student answers exactly 6 questions correctly?
Thank you so much for your help...
Prb(right) = 1/3, prb(wrong) = 2/3
a) you want
rrrrrrwwww
prb(rrrrrrwwww) = (1/3)^6(2/3)^4
= 16/59049
b) you want rrwrwrrwwr or the same as above
c) now the order does not matter, you want to choose 6 out of 10 which is
C(10,6) = 10!/(6!4!) = 210
d) prb(6 out of 10) = 210/59049 = 70/19683
You have a 5-question multiple-choice test. Each question has four choices. You don’t know any of the answers. What is the experimental probability that you will guess exactly three out of five questions correctly?
P(3 correct)= (.25)^3 x (.75)^2 combinations; 5C3= 5!/3!2!= 10
answer is 1/64 x 9/16 x 10 = 90/1024
I have no idea
fake
To answer these questions, we need to understand some basic concepts of probability and combination.
In this case, each question has three possible answers, and the student is randomly guessing the answers. This means the probability of guessing the correct answer for each question is 1/3, and the probability of guessing the incorrect answer is 2/3.
Now let's address each question one by one:
(a) What is the probability that the student will answer the first six questions correctly and the rest incorrectly?
Since the student is guessing randomly, the probability of getting each question correct is 1/3. Since all events are independent (one question's outcome doesn't affect the outcome of another question), we can multiply the probabilities together to get the overall probability.
The probability of answering the first six questions correctly is (1/3)^6, and the probability of answering the remaining four questions incorrectly is (2/3)^4. Therefore, the probability that the student will answer the first six questions correctly and the rest incorrectly is (1/3)^6 * (2/3)^4.
(b) What is the probability that the student will answer the 1st, 2nd, 4th, 6th, 7th, and 10th questions correctly and the rest incorrectly?
Similarly, the probability of guessing the correct answer for each of the mentioned questions is 1/3. The probability of guessing incorrectly for the remaining four questions is 2/3.
Therefore, the probability of answering the 1st, 2nd, 4th, 6th, 7th, and 10th questions correctly and the rest incorrectly is (1/3)^6 * (2/3)^4.
(c) How many ways can the student get exactly 6 correct answers?
To determine this, let's use the concept of combinations. The student needs to choose exactly 6 questions out of 10 to answer correctly. The order doesn't matter in this case.
The number of ways to choose 6 questions out of 10 can be calculated using the combination formula: C(n, r) = n! / (r!(n-r)!)
In this case, n (the total number of questions) is 10, and r (the number of questions to be answered correctly) is 6.
Therefore, the number of ways the student can get exactly 6 correct answers is C(10, 6) = 10! / (6!(10-6)!) = 210 ways.
(d) What is the probability that the student answers exactly 6 questions correctly?
To find the probability, we need to divide the number of favorable outcomes (getting exactly 6 correct answers) by the total number of possible outcomes (answering all 10 questions).
The total number of possible outcomes is 3^10, as there are three choices for each question and a total of 10 questions.
Therefore, the probability of the student answering exactly 6 questions correctly is 210 / (3^10).
I hope this explanation helps clarify how to approach each question.