If there are 4 choices and you randomly choose one, you have a 25% chance that it will be correct, with 75% that it will be wrong. However, if there is one answer that you know is not correct, you can eliminate that one. Then if you choose from the remaining three, you will have increased your chances of getting it right to 33%. That doesn't sound like a whole lot, but it really would help you.
64/256
4/25
The probability of correct true & false question is 1/2 and the probability correct multiple choice (four answer) question is 1/4. We want the probability of correct, correct, and correct. Therefore the probability all 3 questions correct is 1/2 * 1/2 * 1/4 = 1/16.
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
It is 0.0033
64/256
15%? (My math sucks - I probably got that wrong).
4/25
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.
To find the probability of getting at least 6 correct answers on a 10-question multiple-choice exam where each question has 5 choices (with only one correct answer), we can model this situation using the binomial probability formula. The probability of guessing correctly on each question is ( p = \frac{1}{5} ) and incorrectly is ( q = \frac{4}{5} ). We need to calculate the sum of probabilities for getting exactly 6, 7, 8, 9, and 10 correct answers. Using the binomial formula, the probability ( P(X = k) ) for each ( k ) can be computed, and then summed to find ( P(X \geq 6) ). The resulting probability is approximately 0.0163, or 1.63%.
25
The probability of correct true & false question is 1/2 and the probability correct multiple choice (four answer) question is 1/4. We want the probability of correct, correct, and correct. Therefore the probability all 3 questions correct is 1/2 * 1/2 * 1/4 = 1/16.
In a Binomial distribution, if a student randomly guesses on multiple-choice questions with 5 possible choices, the probability of selecting the correct answer is ( p = \frac{1}{5} ) and the probability of selecting an incorrect answer is ( q = 1 - p = \frac{4}{5} ). The expected score for a student guessing on ( n ) questions is calculated as ( E(X) = n \cdot p ). To ensure that a student who randomly guesses has an expected score of 0, the number of questions ( n ) must be set to 0, or alternatively, the scoring system must be adjusted so that the expected value of scoring remains zero, such as by introducing penalties for incorrect answers.
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
Each guess has a 25% chance of being correct and a 75% chance of being wrong. Guessing right or wrong on one question does not affect the odds on the next one.