What is the probability of answering at least 1 question out of 5 correctly?

Question

Each question has 4 possible choices.
Can't seem to work out the math for this.

Answer 1

It's easier to do this one in reverse:

The odds of getting none right is (3/4)^5 = 243/1024

So, the odds of getting at least one right is 1 - 243/1024. That's about 76%

Answer 2

The probability of answering 1 question correctly is 1/4. Assuming you are guessing totally randomly. Because there are 4 possible choices. Your chances don't get better or worse just because you are answering 5 questions. So since it says At Least one, the probablility is 1/4.

Answer 3

If here are five questions with four answers each, there are 20 possible answers (5 x 4 = 20). If you answer (guess?) all five questions, you will take five chances to hit 20 answers. 5/20 = 0.25 x!00% = 25%. Neglecting the fact that there may be one or more correct answers to each question, there are 5 correct answers to 20 answers or 25% of the answers are correct.

Answer 4

First you must be certain how you are asking the question
(A)The probibility of answering 1 question out of 5 correctly is much different from
(B)the probability fo answering at least one question out of five correctly.

The probability of answering 1 correctly is
P(1) (for ease of writing)

For Question (A) you simply need to calculate P(1)

For this problem
p=(1/4) (1 correct answer for each problem)
q=(3/4) (3 incorrect answers for each problem)
n=5
k=1

p(k success of n trials)=
n!/((n-k)!k!)*p^k*q^(n-k)
p(1 success of 5 trials)=
5!/((5-1)!1!)*(1/4)^1*(3/4)^(4) = 405/1024

But for Question (B) you must Calculate P(1)+P(2)+P(3)+P(4)+P(5)

This is based out of binomial probability http://en.wikipedia.org/wiki/Binomial_probability#Example

If you refer to the article and see the "N over K in parentheses" that is simplified to n!/((n-k)!k!)

p(k success of n trials)=
n!/((n-k)!k!)*p^k*q^(n-k)

p=probability of a success
q=probability of a failure q=1-p.
n=number of trials
k is the number of successes

For this problem
p=(1/4)
(1 correct answer for each problem)
q=(3/4)
(3 incorrect answers for each problem)
n=5
k=k

p(k success of 5 trials)=
5!/((5-k)!k!)*(1/4)^k*(3/4)^(5-k)

p(1 success of 5 trials)=
5!/((5-1)!1!)*(1/4)^1*(3/4)^(4) =
405/1024

p(2 success of 5 trials)=
5!/((5-2)!2!)*(1/4)^2*(3/4)^(5-2)=
270/1024

p(3 success of 5 trials)=
5!/((5-3)!3!)*(1/4)^3*(3/4)^(5-3)
=90/1024

p(4 success of 5 trials)=
5!/((5-4)!4!)*(1/4)^4*(3/4)^(5-4)
=15/1024

p(5 success of 5 trials)=
5!/((5-5)!5!)*(1/4)^5*(3/4)^(5-5)
=1/1024

Now add them all up for 781/1024
(divide for a decimal)

I apologize for the poor formatting if not everything displays, Yahoo answers seems to be incredibly difficult today. E-mail or IM me and I will send a better detailed one.

Answer 5

16/20 I think

Answer 6

It depends on the person who is answering.