What is the probability of getting 80% or more of the questions correct on a 10-question true-false exam merely by guessing?
probability - Damon, Wednesday, March 6, 2013 at 2:53am
Binomial distribution with p = .5
p(10 right)
= c(10,10) .5^10 (.5)^0 =.5^10 = .000976
p( 9 right)
= c(10,9) .5^9 (.5)^1 = .00976
p(8 right)
= c(10,8) (.5)^8 (.5)^2 = 45*.5^10 = .04394
sum = .0547
I can't understand that way.... can anyone explain or solve it another way
well you can understand it like,
for one qusetion getting right,
the probability is=.5
as there are only two options true and false.
for two qusetion getting right,
the probability is=.5*.5
for three qusetion getting right,
the probability is=.5*.5*.5
.................................................................
for eight qusetion getting right,
the probability is=(.5)^8
for exactly eight question getting right you have to make 2 questions wrong.
for two qusetion getting wrong,
the probability is=(.5)^2
and you can select 8 questions out of 10 at a probability=C(10,10)*(.5)^8*(.5)^2
.
.
.
thus you can calculate probability for exactly 9 qstn and 10 qstn right.
sum them and u'ld get the ans.
thanx :) i got it
To find the probability of getting 80% or more of the questions correct on a 10-question true-false exam merely by guessing, we can use the binomial distribution formula.
First, let's define the variables:
n = total number of questions = 10
k = number of questions answered correctly to achieve 80% or more correct = 8, 9, or 10
p = probability of getting a question correct through guessing = 0.5 (since it's a true-false exam)
Now, let's calculate the probabilities:
P(10 right) = C(10,10) * (0.5)^10 * (0.5)^(10-10)
= 1 * (0.5)^10 * 1
= 0.000976
P(9 right) = C(10,9) * (0.5)^9 * (0.5)^(10-9)
= 10 * (0.5)^9 * (0.5)^1
= 0.00976
P(8 right) = C(10,8) * (0.5)^8 * (0.5)^(10-8)
= 45 * (0.5)^8 * (0.5)^2
= 0.04394
Now, sum up the probabilities:
P(80% or more correct) = P(10 right) + P(9 right) + P(8 right)
= 0.000976 + 0.00976 + 0.04394
= 0.054676
So, the probability of getting 80% or more of the questions correct on a 10-question true-false exam merely by guessing is approximately 0.0547 or 5.47%.
To find the probability of getting 80% or more of the questions correct on a 10-question true-false exam merely by guessing, we can use the binomial distribution.
The binomial distribution is used when there are two possible outcomes (in this case, correct or incorrect) and each outcome has a fixed probability (in this case, 50% since the student is guessing). We can calculate the probability of each possible outcome and sum them up.
To start, let's calculate the probability of getting exactly 10 questions right. The formula for this is:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
where:
- P(X = k) is the probability of getting exactly k questions right
- C(n, k) is the number of ways to choose k items from a set of n items (binomial coefficient)
- p is the probability of getting one question right
- n is the total number of questions
In this case, p = 0.5 (since the student is guessing), n = 10, and k = 10. Plugging these values into the formula, we get:
P(X = 10) = C(10, 10) * (0.5)^10 * (0.5)^0
= 1 * (0.5)^10 * 1
= 0.000976
So the probability of getting exactly 10 questions right is 0.000976.
We can repeat this calculation for getting exactly 9 questions right, 8 questions right, and so on. Finally, we sum up the probabilities of all these outcomes.
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)
P(X = 8) can be calculated as:
P(X = 8) = C(10, 8) * (0.5)^8 * (0.5)^2
= 45 * (0.5)^10
= 0.04394
Similarly, we can calculate P(X = 9) = 0.00976.
Summing up all these probabilities, we get:
P(X ≥ 8) = 0.04394 + 0.00976 + 0.000976
= 0.05467
Therefore, the probability of getting 80% or more of the questions correct on a 10-question true-false exam merely by guessing is approximately 0.05467 or 5.467%.