A biology quiz consists of 10 multiple -choice questions. 7 must be answered correctly to receive a passing grade. If each question has 4 possible answers, of which only one is correct, what is the probability that a student who guesses at random on each question will pass the examination? (round to four decimal places)
To find the probability that a student will pass the examination by guessing at random, we need to determine the probability of answering at least 7 questions correctly out of 10.
The probability of answering a question correctly by guessing is 1/4, since there is only one correct answer out of four possible options.
To calculate the probability of answering a specific number of questions correctly out of 10, we can 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) is the number of combinations of n items taken k at a time,
p is the probability of success on a single trial, and
n is the number of trials.
In this case, n = 10 (number of questions), k ≥ 7 (number of questions answered correctly), and p = 1/4 (probability of guessing correctly).
We need to calculate the sum of probabilities for k = 7, 8, 9, and 10 questions answered correctly:
P(X≥7) = P(X=7) + P(X=8) + P(X=9) + P(X=10)
Using the formula above and substituting the values, we can calculate:
P(X=7) = C(10,7) * (1/4)^7 * (3/4)^(10-7)
P(X=8) = C(10,8) * (1/4)^8 * (3/4)^(10-8)
P(X=9) = C(10,9) * (1/4)^9 * (3/4)^(10-9)
P(X=10) = C(10,10) * (1/4)^10 * (3/4)^(10-10)
Calculating these values:
P(X=7) = 10C7 * (1/4)^7 * (3/4)^(10-7) = 120 * (1/4)^7 * (3/4)^3
P(X=8) = 10C8 * (1/4)^8 * (3/4)^(10-8) = 45 * (1/4)^8 * (3/4)^2
P(X=9) = 10C9 * (1/4)^9 * (3/4)^(10-9) = 10 * (1/4)^9 * (3/4)^1
P(X=10) = 10C10 * (1/4)^10 * (3/4)^(10-10) = (1/4)^10 * (3/4)^0
Adding these probabilities:
P(X≥7) = P(X=7) + P(X=8) + P(X=9) + P(X=10)
P(X≥7) = 120 * (1/4)^7 * (3/4)^3 + 45 * (1/4)^8 * (3/4)^2 + 10 * (1/4)^9 * (3/4)^1 + (1/4)^10
Evaluating this expression will give us the probability of passing the examination by guessing at random.
To solve this problem, we can use the concept of binomial probability.
The probability of guessing the correct answer to a single question is 1/4 since there are 4 possible answers and only one of them is correct. Similarly, the probability of guessing the incorrect answer is 3/4.
In order to pass the exam, the student needs to answer at least 7 questions correctly. We can calculate the probability of getting exactly 7, 8, 9, or 10 questions correct, and sum them up to get the overall probability of passing.
Let's calculate the probabilities for each case:
1. Getting exactly 7 questions correct:
The student needs to guess correctly on 7 questions and incorrectly on 3 questions. The order of correct and incorrect answers doesn't matter, so we can use combinatorial notation (nCr). The probability would be: (10C7) * (1/4)^7 * (3/4)^3.
2. Getting exactly 8 questions correct:
The student needs to guess correctly on 8 questions and incorrectly on 2 questions. The probability would be: (10C8) * (1/4)^8 * (3/4)^2.
3. Getting exactly 9 questions correct:
The student needs to guess correctly on 9 questions and incorrectly on 1 question. The probability would be: (10C9) * (1/4)^9 * (3/4)^1.
4. Getting exactly 10 questions correct:
The student needs to guess correctly on all 10 questions. The probability would be: (10C10) * (1/4)^10 * (3/4)^0.
Now, we can calculate the sum of these probabilities to get the overall probability of passing:
P(passing) = P(exactly 7) + P(exactly 8) + P(exactly 9) + P(exactly 10)
Let's compute this probability:
P(exactly 7) = (10C7) * (1/4)^7 * (3/4)^3
P(exactly 8) = (10C8) * (1/4)^8 * (3/4)^2
P(exactly 9) = (10C9) * (1/4)^9 * (3/4)^1
P(exactly 10) = (10C10) * (1/4)^10 * (3/4)^0
P(passing) = P(exactly 7) + P(exactly 8) + P(exactly 9) + P(exactly 10)
Calculate each probability and then sum them up to find the overall probability. Round the answer to four decimal places.
I will calculate the probabilities and sum them up for you.