a student randomly guesses at five multiple-choice questions, each question has five possible choices (suppose the choice for each question is independent): a) Find the probability that the student gets exactly three correct. b) What is the probability that the student gets all correct? c) What is the expected number of correct questions? d) Find the probability that the student gets at least the expected number of correct questions ?
To find the answers to the given questions, we need to understand the concept of probability and apply it to this scenario.
Let's break down each question step by step:
a) Find the probability that the student gets exactly three correct.
To solve this, we can use the binomial probability formula. In this case, the student has a 1/5 chance of guessing a correct answer and a 4/5 chance of guessing incorrectly for each question.
The formula for the probability of getting exactly k successes in n independent trials, each with probability of success p, is:
P(X = k) = (nCk) * (p^k) * ((1-p)^(n-k))
Here,
n = number of trials = 5 (as the student guesses 5 questions)
k = number of successes (questions answered correctly) = 3
p = probability of success (guessing correctly) = 1/5
Plugging in these values, the probability of getting exactly three correct answers is:
P(X = 3) = (5C3) * ((1/5)^3) * ((4/5)^(5-3))
Calculating this, we get:
P(X = 3) = (10) * (1/125) * (16/25) ≈ 0.2048
Therefore, the probability that the student gets exactly three correct answers is approximately 0.2048.
b) What is the probability that the student gets all correct?
Since the student is guessing randomly, the probability of getting all answers correct is:
P(X = 5) = (5C5) * (1/5)^5 * (4/5)^(5-5)
= 1 * (1/3125) * 1
= 1/3125
≈ 0.00032
Therefore, the probability that the student gets all correct answers is approximately 0.00032.
c) What is the expected number of correct questions?
To calculate the expected number of correct questions, we need to multiply the probability of getting each possible number of correct questions by the number of correct questions and sum them up.
The expected value (E(X)) formula for a discrete random variable is:
E(X) = Σ(x * P(X = x))
Here, x represents the number of correct questions, and P(X = x) represents the probability of getting x correct questions.
For this scenario:
E(X) = (0 * P(X = 0)) + (1 * P(X = 1)) + (2 * P(X = 2)) + (3 * P(X = 3)) + (4 * P(X = 4)) + (5 * P(X = 5))
Using the probabilities calculated from part a:
E(X) = (0 * 1/3125) + (1 * 60/3125) + (2 * 120/3125) + (3 * 60/3125) + (4 * 10/3125) + (5 * 1/3125)
= 0 + 0.0192 + 0.0768 + 0.096 + 0.0128 + 0.00032
≈ 0.20528
Therefore, the expected number of correct questions is approximately 0.20528.
d) Find the probability that the student gets at least the expected number of correct questions.
To find the probability that the student gets at least the expected number of correct questions, we need to calculate the sum of the probabilities for getting the expected number of correct questions and any greater number.
P(X ≥ E(X)) = P(X = 3) + P(X = 4) + P(X = 5)
Using the probabilities calculated from part a:
P(X ≥ E(X)) = 0.2048 + 0.0128 + 0.00032
≈ 0.21892
Therefore, the probability that the student gets at least the expected number of correct questions is approximately 0.21892.