# Need some help with basic probability notation.

• Feb 21st 2011, 05:36 PM
pantsaregood
Need some help with basic probability notation.
"A multiple choice test consists of four questions, each with four possible answers, only one of which is correct. You randomly guess each answer. What is the probability of:

A: Getting them all right
B: Getting at least one right"

A is (1/4)^4, which is 1/16. B is (1/4)^1, which is 1/4. I don't know what "proper" notation for this is, though.
• Feb 21st 2011, 05:59 PM
harish21
Part A.

\$\displaystyle \bigg(\dfrac{1}{4}\bigg)^4 \neq \dfrac{1}{16}\$

\$\displaystyle \bigg(\dfrac{1}{4}\bigg)^4=\dfrac{1}{256}\$

At least one correct answer means:

\$\displaystyle P(X \geq 1) = P(X=1)+P(X=2)+P(X=3)+P(X=4)\$
• Feb 21st 2011, 06:31 PM
pantsaregood
Well, that's embarrassing. I was thinking 2^4 instead of 4^4. Three Calculus courses, Ordinary Differential Equations, and Linear Algebra down, and I apparently can't work with exponents.

Anyway, I understand that part A is 1/256. I just don't know the proper notation for this; I am in a Calculus based Statistics university course, and my professor is very anal about notation.

I'm not following with B. The chance of getting all four wrong is (3/4)^4. 1-(3/4)^4 would mean you didn't get all four wrong, meaning at least one is correct. Is that correct? What sort of notation is correct for this?
• Feb 21st 2011, 06:52 PM
harish21
Quote:

Originally Posted by pantsaregood
Well, that's embarrassing. I was thinking 2^4 instead of 4^4. Three Calculus courses, Ordinary Differential Equations, and Linear Algebra down, and I apparently can't work with exponents.

Human beings make mistakes. So lets not worry about that!

Quote:

I'm not following with B. The chance of getting all four wrong is (3/4)^4. 1-(3/4)^4 would mean you didn't get all four wrong, meaning at least one is correct. Is that correct? What sort of notation is correct for this?
Yes That is correct. You can conclude that the probability of getting at least one answer correct is: \$\displaystyle P(X \geq 1) = [1-P(X=0)]= 1-\bigg(\dfrac{3}{4}\bigg)^4=0.6835\$

And for part A, the probability \$\displaystyle P(X=4)=\bigg(\dfrac{1}{4}\bigg)^4=\dfrac{1}{256}=0 .0039\$

You can also show the use of the binomial distribution in these problems. I assume you have used binomial distribution to get these answers.

Is that the notation you are talking about?