# Math Help - Probability Question (binomial distribution)

1. ## Probability Question (binomial distribution)

In history class, Colin and Diana both write a multiple choice quiz. there are ten questions. Each question has five possible answers.
What is the probability that
a) Colin will pass the test if he guesses an answer to each question?
b) Diana will pass the test if she studies so that she has a 75% chance of answering each question correctly?

a) This is what i did:
I used binomial distribution (i think some part of this is wrong though)
n=10 p=.5 q=.5 3 out of 5 questions right is passing = .6
10C5(.6)^5(.4)^5
= 0.2006

the answer in the back of the book is 0.033

b) p=.75 q=.25 n=10
10C5(.75)^5(.25)^5
=0.0583

the answer in the back of the book is 0.980

I know the way i did this makes no sense at all because the probability of diana passing should be greater then that of Colin but this is the only way i can think of doing this.
I feel i'm messing up with the exponents.. but to pass on a 10 question test wouldn't you at least have to get 5 questions right ..?
sooo confused...

2. Originally Posted by tmas
In history class, Colin and Diana both write a multiple choice quiz. there are ten questions. Each question has five possible answers.
What is the probability that
a) Colin will pass the test if he guesses an answer to each question?
b) Diana will pass the test if she studies so that she has a 75% chance of answering each question correctly?

a) This is what i did:
I used binomial distribution (i think some part of this is wrong though)
n=10 p=.5 q=.5 3 out of 5 questions right is passing = .6
10C5(.6)^5(.4)^5
= 0.2006

the answer in the back of the book is 0.033

b) p=.75 q=.25 n=10
10C5(.75)^5(.25)^5
=0.0583

the answer in the back of the book is 0.980

I know the way i did this makes no sense at all because the probability of diana passing should be greater then that of Colin but this is the only way i can think of doing this.
I feel i'm messing up with the exponents.. but to pass on a 10 question test wouldn't you at least have to get 5 questions right ..?
sooo confused...
Fist you have written or plugged into the binomial distribution incorrectly

$\displaystyle P(x)=\binom{n}{x}p^{x}(1-p)^{n-x}$

This gives the probability that exactly x events occurred.

So for our good friend the guessing Colin since there are 5 choices and he picks at random his probability of success is $p=\frac{1}{5}=.2$

You have also said for Colin to pass he must get at least 5 questions correct. So this means he will pass if he gets 5,6,7,8,9, or 10 questions correct. So we need to find each of these probabilities and sum them up. So here is the binomial theorem

$\displaystyle P(x)=\binom{10}{x}(0.2)^{x}(0.8)^{10-x}$

So our answer is $\displaystyle P(5)+P(6)+...+P(9)+P(10)=\sum_{x=5}^{10}P(x)=0.032 79$

This should get you started.