Thread: Binomial distribution related to rolling dies

1. Binomial distribution related to rolling dies

A multiple-choice test consists of eight questions and three answers to each question (of which only one is correct). If a student answers each question by rolling a balanced die and checking the first answer if he gets 1 or 2, the second answer if he gets 3 or 4, and the third answer if he gets 5 or 6, what is the probability that he will get exactly four correct answers?

This is a pretty straightforward binomial question, the probability of success (ie, getting one question correct) is given by 1/3, then we can directly plug it into the binomial probability distribution function to obtain the probabilities, however I'm wondering how the question would change if it was stated like this:

What if he chooses the second answer if he gets a 3 or 4 or 5, chooses the first answer if he gets a 1 or 2 and chooses answer 3 if he gets a 6?

How would I work out the probability of getting one question correct now?

Thanks

2. Re: Binomial distribution related to rolling dies

Originally Posted by usagi_killer
What if he chooses the second answer if he gets a 3 or 4 or 5, chooses the first answer if he gets a 1 or 2 and chooses answer 3 if he gets a 6?

How would I work out the probability of getting one question correct now?

Thanks
This would depend on which answer was correct in each question. Even if you knew this it would it would be a mess to work out.

btw the plural of die is dice.

3. Re: Binomial distribution related to rolling dies

Ahh thanks, I thought so as well

lol I typo'd the title XD

4. Re: Binomial distribution related to rolling dies

Actually, it doesn't matter how he makes his random choice, the probability that he will get the question right is still 1/3, provided that all the answers are equally likely to be correct, and the guess and correct answer are independent events.

Let's say $\displaystyle A_i$ means answer i is correct and $\displaystyle G_i$ means the guess is answer i.

Then the probability that he guesses the answer correctly is
$\displaystyle \sum_{i=1}^3 P(A_i G_i) = \sum_{i=1}^3 P(A_i) P(G_i) = \sum_{i=1}^3 (1/3) P(G_i) = (1/3) \sum_{i=1}^3 P(G_i ) = (1/3) \cdot 1 = 1/3$