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lesson 10-6 practice a theoretical probability find the theoretical probability of each outcome. 1. flipping one coin and having it land tails up 2. randomly choosing a yellow marble from a bag of 3 yellow marbles and 7 blue marbles 3. rolling a 1 on a number cube 4. randomly choosing the letter a from the letters in math 5. the probability it will rain is 10%. what is the probability it will not rain? 6. the probability of choosing a red marble from a bag is \\(\frac{3}{4}\\). what is the probability of not choosing a red marble? 7. a spinner has red, green, and blue. the probability of spinning red is 0.2 and the probability of spinning blue is 0.3. what is the probability of spinning a green? 8. the probability of winning first place in a contest is 5%. what is the probability of not winning first place? 9. the odds in favor of winning a contest are 1:4. a. what is the total number of possible outcomes? b. how many ways can a person win? c. what is the probability that a person wins? 10. the odds against a spinner landing on green are 7:2. a. what is the total number of possible outcomes? b. how many ways can the spinner not land on green? c. what is the probability of the spinner not landing on green? 11. the probability of choosing a green marble from a bag is \\(\frac{3}{5}\\). a. how many ways can green be chosen? b. how many ways can green not be chosen? c. what are the odds in favor of choosing a green marble? 12. the table shows how many of each letter are in a bag. find the following. a. \\(p(a)\\) b. \\(p(\text{not } b)\\) c. odds in favor of c \

$$\begin{tabular}{|c|c|} \\hline letter & how many in bag \\\\ \\hline a & 6 \\\\ \\hline b & 4 \\\\ \\hline c & 5 \\\\ \\hline \\end{tabular}$$

Explanation:

Step1: Solve problem 1 (flipping a coin for tails)

A fair coin has 2 possible outcomes: heads (H) or tails (T). The theoretical probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. For landing tails, the number of favorable outcomes is 1 (tails), and the total number of outcomes is 2 (heads or tails). So the probability \( P(\text{tails})=\frac{1}{2} \).

Step2: Solve problem 2 (choosing a yellow marble)

The bag has 3 yellow marbles and 7 blue marbles. So the total number of marbles is \( 3 + 7=10 \). The number of favorable outcomes (yellow marbles) is 3. Thus, the probability \( P(\text{yellow})=\frac{3}{10} \).

Step3: Solve problem 3 (rolling a 1 on a number cube)

A standard number cube (die) has 6 faces, numbered 1 - 6. The number of favorable outcomes (rolling a 1) is 1, and the total number of outcomes is 6. So \( P(1)=\frac{1}{6} \).

Step4: Solve problem 4 (choosing letter A from MATH)

The word MATH has 4 letters: M, A, T, H. The number of favorable outcomes (letter A) is 1, and the total number of letters is 4. So \( P(A)=\frac{1}{4} \).

Step5: Solve problem 5 (probability of not raining)

The probability of an event and its complement (the event not happening) sum to 1. If the probability of rain is 10% or \( 0.1 \), then the probability of not raining is \( 1 - 0.1 = 0.9 \) or 90%.

Step6: Solve problem 6 (probability of not choosing a red marble)

Let \( P(\text{red})=\frac{3}{4} \). The probability of the complement (not red) is \( 1 - \frac{3}{4}=\frac{1}{4} \).

Step7: Solve problem 7 (probability of spinning green)

The sum of the probabilities of all possible outcomes in a sample space is 1. Let \( P(\text{red}) = 0.2 \), \( P(\text{blue})=0.3 \), and \( P(\text{green})=x \). Then \( 0.2+0.3 + x=1 \). Solving for \( x \), we get \( x = 1-(0.2 + 0.3)=0.5 \).

Step8: Solve problem 8 (probability of not winning first place)

The probability of winning first place is 5% or \( 0.05 \). The probability of not winning is \( 1 - 0.05 = 0.95 \) or 95%.

Step9: Solve problem 9 (odds in favor 1:4)

Part a: Total number of possible outcomes

Odds in favor of an event are given as \( \text{number of favorable}:\text{number of unfavorable} \). So if odds in favor are 1:4, the total number of outcomes is \( 1 + 4=5 \).

Part b: Number of ways to win

From the odds 1:4, the number of favorable outcomes (winning) is 1.

Part c: Probability of winning

Probability is \( \frac{\text{favorable}}{\text{total}}=\frac{1}{5} \).

Step10: Solve problem 10 (odds against 7:2)

Part a: Total number of possible outcomes

Odds against an event are \( \text{number of unfavorable}:\text{number of favorable} \). So if odds against green are 7:2, total outcomes are \( 7 + 2 = 9 \).

Part b: Number of ways to not land on green

From odds against 7:2, the number of unfavorable outcomes (not green) is 7.

Part c: Probability of not landing on green

Probability is \( \frac{\text{unfavorable}}{\text{total}}=\frac{7}{9} \).

Step11: Solve problem 11 (probability of green marble \( \frac{3}{5} \))

Part a: Number of ways to choose green

If \( P(\text{green})=\frac{3}{5} \), we can think of the number of green marbles as 3 (assuming the denominator 5 represents total parts, so green is 3 parts).

Part b: Number of ways to not choose green

The total number of parts is 5, so not green is \( 5 - 3 = 2 \) parts.

Part c: Odds in favor of green

Odds in favor are \( \text{favorable}:\text{unfavorable}=3:2 \) or \( \frac{3}{2} \).

Step12: Solve problem 12 (letters in a bag: A:6, B:4, C:5)

Part a: \( P(A) \)

Total number of letters: \( 6 + 4+5 = 15 \). Number of A's is 6. So \( P(A)=\frac{6}{15}=\frac{2}{5} \).

Part b: \( P(\text{not B}) \)

Number of B's is 4, so number of not B's is \( 15 - 4 = 11 \). Thus, \( P(\text{not B})=\frac{11}{15} \).

Part c: Odds in favor of C

Number of C's (favorable) is 5, number of not C's is \( 15 - 5 = 10 \). Odds in favor are \( 5:10 = 1:2 \) or \( \frac{1}{2} \).

Answers (for each problem):

1. \( \boldsymbol{\frac{1}{2}} \)
2. \( \boldsymbol{\frac{3}{10}} \)
3. \( \boldsymbol{\frac{1}{6}} \)
4. \( \boldsymbol{\frac{1}{4}} \)
5. \( \boldsymbol{0.9} \) (or 90%)
6. \( \boldsymbol{\frac{1}{4}} \)
7. \( \boldsymbol{0.5} \)
8. \( \boldsymbol{0.95} \) (or 95%)
9. a. \( \boldsymbol{5} \); b. \( \boldsymbol{1} \); c. \( \boldsymbol{\frac{1}{5}} \)
10. a. \( \boldsymbol{9} \); b. \( \boldsymbol{7} \); c. \( \boldsymbol{\frac{7}{9}} \)
11. a. \( \boldsymbol{3} \); b. \( \boldsymbol{2} \); c. \( \boldsymbol{\frac{3}{2}} \) (or \( 3:2 \))
12. a. \( \boldsymbol{\frac{2}{5}} \); b. \( \boldsymbol{\frac{11}{15}} \); c. \( \boldsymbol{\frac{1}{2}} \) (or \( 1:2 \))

Answer:

s (for each problem):

1. \( \boldsymbol{\frac{1}{2}} \)
2. \( \boldsymbol{\frac{3}{10}} \)
3. \( \boldsymbol{\frac{1}{6}} \)
4. \( \boldsymbol{\frac{1}{4}} \)
5. \( \boldsymbol{0.9} \) (or 90%)
6. \( \boldsymbol{\frac{1}{4}} \)
7. \( \boldsymbol{0.5} \)
8. \( \boldsymbol{0.95} \) (or 95%)
9. a. \( \boldsymbol{5} \); b. \( \boldsymbol{1} \); c. \( \boldsymbol{\frac{1}{5}} \)
10. a. \( \boldsymbol{9} \); b. \( \boldsymbol{7} \); c. \( \boldsymbol{\frac{7}{9}} \)
11. a. \( \boldsymbol{3} \); b. \( \boldsymbol{2} \); c. \( \boldsymbol{\frac{3}{2}} \) (or \( 3:2 \))
12. a. \( \boldsymbol{\frac{2}{5}} \); b. \( \boldsymbol{\frac{11}{15}} \); c. \( \boldsymbol{\frac{1}{2}} \) (or \( 1:2 \))