For each question the student has a 50% chance of answering right.
We can express this as a fraction of 1 (1 being 100%): 1/2, i.e. 0.5
There are 5 questions, so the answer will be:
0.5*0.5*0.5*0.5*0.5=0.5^5=0.03125
or 3.125%
What is the probability of what?Guessing them all correctly?Getting half of the correct?Getting them all wrong?PLEASE be specific with your questions if you want WikiAnswers to help.
This is abinomial distribution; number of trials (n) is 5, probability of success (p) is 1/4 or 0.25. With this information you can go to a Binomial Distribution Table and find the solution. Within the section of values for n=5 and p=.25, read from the section the probability of 4 which is 0.0146 (see related link for table).
5 out of 10
The probability of getting at least 1 answer correct = 1 - Probability of getting all answers correct.So in your case it for be P(at least 1 answer correct) = 1 - 1/256where 256 is your sample space, |S| = 2^8.
The probability of getting at least 1 correct answer is equal to one minus theprobability of answering all incorrect, this would be;P(atleast 1 correct) =1 - P(allincorrect) =1 - (1/2)8 =1 - 0.00390625 ~~ 0.9961 ~ 99.61%
In a multiple-choice test with 4 options (a, b, c, d) for each question, the probability of guessing correctly for each question is ( \frac{1}{4} ). If a student guesses on 10 questions, the expected number of correct guesses can be calculated by multiplying the number of questions by the probability of a correct guess: ( 10 \times \frac{1}{4} = 2.5 ). Therefore, the mean expected correct guesses for the student is 2.5.
What is the probability of what?Guessing them all correctly?Getting half of the correct?Getting them all wrong?PLEASE be specific with your questions if you want WikiAnswers to help.
2
It is 0.0033
The probability that the student will pass is; P(pass) = P(10) + P(9) + P(8) = [10C10 + 10C9 + 10C8] / (.5)10 = 56/1024 ~ ~ 0.0547 ~ 5.47% where nCr = n!/[r!(n-r)!]
7/128, or about 5.5% The student has a 1/2 probability of getting each question correct. The probability that he passes is the probability that he gets 10 correct+probability that he gets 9 correct+probability that he gets 8 correct: P(passes)=P(10 right)+P(9 right)+P(8 right)=[(1/2)^10]+[(1/2)^10]*10+[(1/2)^10]*Combinations(10,2)=[(1/2)^10](1+10+45)=56/1024=7/128.
This is abinomial distribution; number of trials (n) is 5, probability of success (p) is 1/4 or 0.25. With this information you can go to a Binomial Distribution Table and find the solution. Within the section of values for n=5 and p=.25, read from the section the probability of 4 which is 0.0146 (see related link for table).
Assuming the questions are answered at random, the probability is 0.000009, approx.
5 out of 10
If there are four possible answers to a question, then a guessed answer would have a probability of 1 in 4. If there are six questions, then the mean number of correct answers would be six times 1 in 4, or 1.5
Since there are only two options for the answer, on average the student will answer half of the answers correctly.
0.05 I think is the answer