# Homework Help: Mathematics of Data Management - Probability distributiono

1. Jan 17, 2012

### zyQUza0e5esy2

1. The problem statement, all variables and given/known data
A spinner has eight equally-sized sectors, numbered 1 through 8. What is the probability that the arrow on the spinner will stop on a prime number?

2. Relevant equations
P(x) = 1 / n, out come of a uniform probability distribution
P(x) = Probability distribution
n = Total # of possible outcomes

3. The attempt at a solution
n = 8, since the total number of possible outcome range from 1 through 8
P(x) = 1/n
Prime numbers are 1, 3, 5, 7
*Note that some textbooks don't include 1 as a prime number

P(x) = 1/n
= 1/8
** This is wrong but that's how it seems like it is suppose to be done

P(x) = 4/8
4 = total number of primes
P(x) = 1/2 or 3/8 ( if you don't consider 1 as a prime number)
** this is the correct answer

The second way of solving it is the correct answer but what i don't understand is that the formula is P(x) = 1/n, why did 1 become a 4, how would i know that i'm suppose to change the 1? and yes i have checked the 1 is not an L or an I...

Last edited: Jan 17, 2012
2. Jan 17, 2012

### lanedance

the probability of any single number coming up is 1/8. As you have 4 (or 3) potential outcomes, the probability is 4/8 (or 3/8)

3. Jan 17, 2012

### zyQUza0e5esy2

But why is it 4/8 or 3/8? the formula is p(x) = 1/n

4. Jan 17, 2012

### lanedance

not sure if I'm understanding you

say you have n, distinct evenly distributed outcomes (labelled 1 to n)

the probability of getting an outcome x is p(x) = 1/n

the probabilty of getting one of m outcomes is m/n

If you want to break it right down, you know
$$p(x)=\frac{1}{8}$$

So
$$p(1)=p(3)=p(5)=p(7)=\frac{1}{8}$$

you also know only a single number can appear at a time, they are mutually exclusive events
$$p(i\cap j)=0$$

then, as the intersection is zero (mutually exclusive)
$$p(1\cup3 \cup 5\cup7)=p(1)+p(3)+p(5)+p(7)= \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}= \frac{4}{8}$$

Last edited: Jan 17, 2012
5. Jan 17, 2012

### zyQUza0e5esy2

My lord! Thank you that is just perfect!!!!!!!!!!!!!<3