Answer to Question #313681 in Statistics and Probability for Kate malona

Question

Samples of 3 cards are drawn from a population of five cards numbered from 1-5.

1. How many are the possible outcomes?

2. What are the possible means?

3. What is the probability of getting 4 as a mean?

4. What is the probability of getting 2 as a mean?

5. What is the probability of getting 3.33 as a mean?

Answer 1

1.    There are $C_{5}^{3}=10$ possible outcomes.

2.    The outcomes are

$\left( 1,2,3 \right) ,\bar{x}=2\\\left( 1,2,4 \right) ,\bar{x}=\frac{7}{3}\\\left( 1,2,5 \right) ,\bar{x}=\frac{8}{3}\\\left( 1,3,4 \right) ,\bar{x}=\frac{8}{3}\\\left( 1,3,5 \right) ,\bar{x}=3\\\left( 1,4,5 \right) ,\bar{x}=\frac{10}{3}\\\left( 2,3,4 \right) ,\bar{x}=3\\\left( 2,3,5 \right) ,\bar{x}=\frac{10}{3}\\\left( 2,4,5 \right) ,\bar{x}=\frac{11}{3}\\\left( 3,4,5 \right) ,\bar{x}=4$

The possible means are $2,\frac{7}{3},\frac{8}{3},3,\frac{10}{3},\frac{11}{3},4$

$3:\\P\left( \bar{x}=4 \right) =P\left( \left( 3,4,5 \right) \right) =\frac{1}{10}\\4:\\P\left( \bar{x}=2 \right) =P\left( \left( 1,2,3 \right) \right) =\frac{1}{10}\\5:\\P\left( \bar{x}=3.33 \right) =P\left( \bar{x}=\frac{10}{3} \right) =P\left( \left( 1,4,5 \right) \right) +P\left( \left( 2,3,5 \right) \right) =\frac{2}{10}=\frac{1}{5}\\$