Exam 2 Prep

1

Consider the random experiment of drawing a simple random sample of
size 2 from the population consisting of 6 adults: Amy, Brian, Caleb, Debbie, Eric, and Faith.

Explain why using classical probability is appropriate for this random experiment
We can create/visualize every possible outcome in the sample space, so classical probability can be easily used.

Construct the sample space for the random experiment. Use the first letter of each name to represent the individual$$ { AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, DE } $$
Explain how to determine the size of the sample space without having to write all possible outcomes explicitly
To calculate the sample space size, we use Newton's binomial $$ |S| = \binom{6}{2} = \frac{6!}{2! (6 - 2)!} = \frac{6 \cdot 5}{2} = 15 $$

Determine the probability of the event that the sample contains only males
M - only males
$M = {B C, B E, C E}$
$P (M) = \frac{| M |}{| S |} = \frac{3}{15} = \frac{1}{5} = 20 %$

Determine the probability of the event that the sample is unbiased
population parameter - 50% male, 50% female
unbiased sample - $\hat{p} = p = 0.5$
$U = {A B, A C, A E, B D, B F, C D, C F, D E, D F}$ $| U | = 9$
$P (U) = \frac{9}{15} = \frac{3}{5} = 0.6 = 60 %$

Determine the complement of the sample that the sample contains no females
Complement - only males
Already calculated - $20 %$

If the order of the selection mattered in forming our samples of size 2, how many different samples could there be
If the order is significant, we calculate the sample space size: $$ |S'| = \overset{\text{first person}}{6} \cdot \overset{\text{second person}}{5} = 30 $$

2

The first unit exam consisted of 8 questions. The first 2 questions were from the first chapter. The last 6 questions were from the second chapter. If I had 4 possible questions for chapter 1 and 14 possible questions for chapter 2, how many different exams could be written from this questions pool while keeping the same format of 2 questions from chapter 1 and 6 questions from chapter 2?

First chapter - $| A | = (\binom{4}{2}) = 6$
Second chapter - $| B | = (\binom{14}{6}) = 3003$
Total - $| T | = | A | \cdot | B | = 6 \cdot 3003 = 18018$

3

A child selected 10 different beads to form a bracelet. Five of the beads have a bluish hue, while the other 5 beads have a greenish hue. the child decides to alternate between the hues. Determine the number of ways the child can construct the bracelet

Starting from blue - $| B | = \overset{B_{1}}{5} \cdot \overset{G_{1}}{5} \cdot \overset{B_{2}}{4} \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1 = 14400$
Starting from green - $| G | = \overset{G_{1}}{5} \cdot \overset{B_{1}}{5} \cdot \overset{G_{2}}{4} \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1 = 14400$
Total - $| B | + | G | = 28800$

4

Consider rolling 3 fair, six-sided dice. Determine the size of the sample space. Determine the probability that the sum of the three values of the dice equals 5

Sample space - $| S | = 6 \cdot 6 \cdot 6 = 6^{3} = 216$
Equal 3 - ${(122), (212), (221), (113), (131), (311)}$ $| A | = 6$
$P (A) = \frac{| A |}{| S |} = \frac{6}{216} = \frac{1}{36}$