In a multiple choice test that contains ten questions with each question having five possible answers, what is the probability that Colin will pass the test if he merely guesses at each question?
Probability of getting the right answer for 1 question = 1 choice out of 5, or p=1/5.
Assuming independence among the 10 questions, probability of getting all 10 right = p10
Probability of passing (getting 6 or more right out of 10) using binomial probabilities
=10C6*p^6q^4 + 10C7*p^7q^3 + 10C8*p^8q^2 + 10C9*p^9q + 10C10p^10
=(10752+1536+144+8+1)/9765625
= 12441/9765625
= 0.00127 approx.
To calculate the probability that Colin will pass the test by merely guessing at each question, we first need to determine the passing grade. Let's assume that Colin needs to answer at least 6 of the 10 questions correctly in order to pass.
Since each question has five possible answers, including only one correct answer, Colin has a 1/5 chance of guessing the correct answer for each question.
To find the probability of Colin guessing exactly 6, 7, 8, 9 or 10 questions correctly, we need to use the binomial probability formula, which is:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting exactly x successes
C(n, x) is the number of combinations of n items taken x at a time (n choose x)
p is the probability of a single success (in this case, 1/5)
n is the number of independent trials (in this case, 10)
Using this formula, we can calculate the individual probabilities for Colin guessing 6, 7, 8, 9, or 10 questions correctly and then sum them up to get the overall probability of passing.
P(passing) = P(6) + P(7) + P(8) + P(9) + P(10)
Calculating each individual probability:
P(6) = C(10, 6) * (1/5)^6 * (4/5)^4
P(7) = C(10, 7) * (1/5)^7 * (4/5)^3
P(8) = C(10, 8) * (1/5)^8 * (4/5)^2
P(9) = C(10, 9) * (1/5)^9 * (4/5)^1
P(10) = C(10, 10) * (1/5)^10 * (4/5)^0
Now, we substitute the values and calculate each probability using a combination (n choose x) formula and arithmetic operations.
P(6) = 210 * (1/5)^6 * (4/5)^4
P(7) = 120 * (1/5)^7 * (4/5)^3
P(8) = 45 * (1/5)^8 * (4/5)^2
P(9) = 10 * (1/5)^9 * (4/5)^1
P(10) = 1 * (1/5)^10 * (4/5)^0
After calculating each probability, we add them together:
P(passing) = P(6) + P(7) + P(8) + P(9) + P(10)
Once you perform these calculations, you will find the probability that Colin will pass the test by merely guessing at each question.