Question

section 3.1 basic concepts of probability and counting 141
finding the probability of an event in exercises 21–24, the probability that an event will not happen is given. find the probability that the event will happen.

1. ( p(e) = 0.95 )
2. ( p(e) = 0.13 )
3. ( p(e) = \frac{3}{4} )
4. ( p(e) = \frac{21}{25} )

using and interpreting concepts
identifying the sample space of a probability experiment in exercises 25–32, identify the sample space of the probability experiment and determine the number of outcomes in the sample space. draw a tree diagram when appropriate.

1. guessing the initial of a student’s middle name
2. guessing a student’s letter grade (a, b, c, d, f) in a class
3. drawing one card from a standard deck of cards
4. identifying a person’s eye color (brown, blue, green, hazel, gray, other) and hair color (black, brown, blonde, red, other)
5. tossing two coins
6. tossing three coins
7. rolling a pair of six - sided dice
8. rolling a six - sided die, tossing two coins, and spinning the fair spinner shown

Explanation:

For Exercises 21-24:

Step1: Recall complement rule

For any event $E$, $P(E) = 1 - P(E')$

Step2: Calculate for Exercise 21

Substitute $P(E')=0.95$
$P(E) = 1 - 0.95$

Step3: Calculate for Exercise 22

Substitute $P(E')=0.13$
$P(E) = 1 - 0.13$

Step4: Calculate for Exercise 23

Substitute $P(E')=\frac{3}{4}$
$P(E) = 1 - \frac{3}{4}$

Step5: Calculate for Exercise 24

Substitute $P(E')=\frac{21}{61}$
$P(E) = 1 - \frac{21}{61}$

1. Sample Space: All 26 letters of the English alphabet. There are 26 outcomes. No tree diagram needed (single categorical outcome set).
2. Sample Space: $\{A, B, C, D, F\}$. There are 5 outcomes. No tree diagram needed (discrete, finite set of grades).
3. Sample Space: All 52 unique cards in a standard deck (13 ranks × 4 suits). There are 52 outcomes. A tree diagram could split first by suit (hearts, diamonds, clubs, spades) then by rank (Ace-King).
4. Sample Space: $\{(\text{brown, black}), (\text{brown, brown}), (\text{brown, blonde}), (\text{brown, red}), (\text{brown, other}), (\text{blue, black}), (\text{blue, brown}), (\text{blue, blonde}), (\text{blue, red}), (\text{blue, other}), (\text{green, black}), (\text{green, brown}), (\text{green, blonde}), (\text{green, red}), (\text{green, other}), (\text{hazel, black}), (\text{hazel, brown}), (\text{hazel, blonde}), (\text{hazel, red}), (\text{hazel, other}), (\text{gray, black}), (\text{gray, brown}), (\text{gray, blonde}), (\text{gray, red}), (\text{gray, other}), (\text{other, black}), (\text{other, brown}), (\text{other, blonde}), (\text{other, red}), (\text{other, other})\}$. There are $6 \times 5 = 30$ outcomes. A tree diagram would have a first branch for eye color (6 options) and second branches for hair color (5 options).
5. Sample Space: $\{HH, HT, TH, TT\}$. There are $2^2=4$ outcomes. A tree diagram has first branch for first coin (H/T), second branch for second coin (H/T).
6. Sample Space: $\{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$. There are $2^3=8$ outcomes. A tree diagram has 3 levels, each with H/T branches.
7. Sample Space: All pairs $(a,b)$ where $a,b \in \{1,2,3,4,5,6\}$. There are $6 \times 6=36$ outcomes. A tree diagram has first branch for die 1 (1-6), second branch for die 2 (1-6).
8. Sample Space: All triples $(d, c, s)$ where $d \in \{1,2,3,4,5,6\}$, $c \in \{H,T\}$, $s$ is the set of outcomes on the spinner (assumed $n$ outcomes, total outcomes = $6 \times 2 \times n$). A tree diagram has 3 levels: die (1-6), coin (H/T), spinner (its distinct outcomes).

Answer:

1. $P(E)=0.05$
2. $P(E)=0.87$
3. $P(E)=\frac{1}{4}$
4. $P(E)=\frac{40}{61}$

---

For Exercises 25-32: