Darby guesses the answers to six questions on the math portion of her college entrance exam. Each question is a true or false question. What is the approximate probability that at least 4 of her answers are correct?
A. 0.23
B. 0.34
C. 0.66
D. 0.78
Explanation please??
The answer is
(n,r)/2^n
Here there are 6 questions, so n=6, and
there are 2^6=64 possible sets of answers.
Since 4 of them are right, the number of combinations with 4 correct out of 6 is
(n,r)=(6,4)=6*5/(1*2)=15
What would be the probability that 4 out of six are correct if the answers are completely random?
To find the approximate probability that at least 4 of Darby's answers are correct, we can use the binomial probability formula. Since Darby has six questions and each question has two possible outcomes (true or false), we can model this situation as a binomial distribution.
The probability of success (p) in this case is 0.5, since there is an equal chance of answering each question correctly or incorrectly.
Now, we want to find the probability of getting at least 4 correct answers out of the 6 questions. This means we need to calculate the probability of getting exactly 4, 5, or 6 correct answers and add them together.
To calculate the probability of getting exactly k successes out of n trials, we use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of getting exactly k successes.
- C(n, k) represents the binomial coefficient, which is the number of ways to choose k successes from n trials, given by n! / (k!(n-k)!).
- p is the probability of success on a single trial.
- (1-p) is the probability of failure on a single trial.
- n is the number of trials or questions.
Using this formula, we can calculate the probabilities of getting exactly 4, 5, and 6 correct answers and then add them together:
P(X >= 4) = P(X = 4) + P(X = 5) + P(X = 6)
To find these individual probabilities, we can substitute the values into the binomial probability formula and calculate:
P(X = 4) = C(6, 4) * (0.5)^4 * (1-0.5)^(6-4) = 15 * 0.0625 * 0.25 = 0.09375
P(X = 5) = C(6, 5) * (0.5)^5 * (1-0.5)^(6-5) = 6 * 0.03125 * 0.5 = 0.09375
P(X = 6) = C(6, 6) * (0.5)^6 * (1-0.5)^(6-6) = 1 * 0.015625 * 1 = 0.015625
Now, we can add these individual probabilities together to get the probability of at least 4 correct answers:
P(X >= 4) = 0.09375 + 0.09375 + 0.015625 ≈ 0.203125
Therefore, the approximate probability that at least 4 of Darby's answers are correct is 0.20, which is not one of the options given. None of the answer choices (A, B, C, D) matches this probability.