# Bernoulli/Binomial Distributions

• Oct 18th 2008, 09:23 PM
Chris L T521
Bernoulli/Binomial Distributions
There is this one part of a problem I'm having trouble setting up...

Quote:

On a six-question multiple-choice test there are five possible answers for each question, of which one is correct (C) and four are incorrect (I). If a student guesses randomly and independently, find the probability of
For this, I let $X=1$ if answer is correct, and $X=0$ if the answer is incorrect.

Quote:

(a) Being correct only on questions 1 and 4 (i.e. C, I, I, C, I, I)
I'm not quite sure on how to set this part up, any hints would be appreciated.

Quote:

(b) Being correct on two questions
I think this would be $P\left(X=2\right)=\binom62\left(\frac{1}{5}\right) ^2\left(\frac{4}{5}\right)^4$...but I'm not quite sure.

I'd appreciate any input! (Nod)

--Chris
• Oct 18th 2008, 10:14 PM
mr fantastic
Quote:

Originally Posted by Chris L T521
There is this one part of a problem I'm having trouble setting up...

On a six-question multiple-choice test there are five possible answers for each question, of which one is correct (C) and four are incorrect (I). If a student guesses randomly and independently, find the probability of

(a) Being correct only on questions 1 and 4 (i.e. C, I, I, C, I, I)

For this, I let $X=1$ if answer is correct, and $X=0$ if the answer is incorrect.

I'm not quite sure on how to set this part up, any hints would be appreciated.

(b) Being correct on two questions
I think this would be $P\left(X=2\right)=\binom62\left(\frac{1}{5}\right) ^2\left(\frac{4}{5}\right)^4$...but I'm not quite sure.

I'd appreciate any input! (Nod)

--Chris

(b) is correct.

For (a), it's simply $\left( \frac{1}{5}\right) \, \left( \frac{4}{5}\right)^2 \, \left( \frac{1}{5}\right) \, \left( \frac{4}{5}\right)^2 = \left( \frac{1}{5}\right)^2 \, \left( \frac{4}{5}\right)^4$.