1. ## Multiple choice questions!

I need help in this probability problems.
This seems like simple, and I tried using the normal probability rule but I was told that this is to be done my binomial, which makes me stuck.

On a given multiple choice test, there are 4 questions, each with 4 possible answers. Say you take the test by guessing randomly on each question.
1. How many questions would you expect to get right?
2. What’s the probability that you get all four questions correct?
3. What’s the probability that you get at least 2 questions correct?

Thanks a lot!

2. Hello,

Originally Posted by Vedicmaths
I need help in this probability problems.
This seems like simple, and I tried using the normal probability rule but I was told that this is to be done my binomial, which makes me stuck.

On a given multiple choice test, there are 4 questions, each with 4 possible answers. Say you take the test by guessing randomly on each question.
Yep, binomial.

Resume : 4 questions -> n=4. 4 possible answers -> p=1/4
Thus this is a binomial distrubition X ~ B (4,1/4)

1. How many questions would you expect to get right?
That is to say the mean of the variable. E(X)=np=1 (Binomial distribution - Wikipedia, the free encyclopedia)

2. What’s the probability that you get all four questions correct?
P(X=4)=^4C_4 * p^4 * (1-p)^{4-4}=p^4=1/4^4=1/64

3. What’s the probability that you get at least 2 questions correct?
P(X >= 2)=1-P(X < 2)=1-P(X=0)-P(X=1)

Try this one

3. Originally Posted by Vedicmaths
I need help in this probability problems.
This seems like simple, and I tried using the normal probability rule but I was told that this is to be done my binomial, which makes me stuck.

On a given multiple choice test, there are 4 questions, each with 4 possible answers. Say you take the test by guessing randomly on each question.
1. How many questions would you expect to get right?
2. What’s the probability that you get all four questions correct?
3. What’s the probability that you get at least 2 questions correct?

Thanks a lot!
1. Sorry, I have not done expected values yet.

2. {4 \choose 4} \left( \frac{1}{4} \right) ^{4} \left( \frac{3}{4} \right) ^{0}

3. {4 \choose 2} \left( \frac{1}{4} \right) ^{2} \left( \frac{3}{4} \right) ^{2} + {4 \choose 3} \left( \frac{1}{4} \right) ^{3} \left( \frac{3}{4} \right) ^{1} + {4 \choose 4} \left( \frac{1}{4} \right) ^{4} \left( \frac{3}{4} \right) ^{0}

4. Originally Posted by Moo
Hello,

Yep, binomial.

Resume : 4 questions -> n=4. 4 possible answers -> p=1/4
Thus this is a binomial distrubition X ~ B (4,1/4)

E(X)=np=1

P(X=4)=^4C_4 * p^4 * (1-p)^{4-4}=p^4=1/16

It must be 1/64? Since 4^4 = 64 and not 16?

P(X >= 2)=1-P(X < 2)=1-P(X=0)-P(X=1)

Try this one
Responses in red.