Question

algebra 2 p1 introduction to probability cw
theoretical probability $=\frac{\\# \text{ favorable outcomes}}{\text{total } \\# \text{ outcomes}}$
using the 2 - dice sample space, find the theoretical probability for the following events:

1. p (rolling a sum of 10)
2. p (rolling doubles)
3. p (rolling a sum divisible by 3)
4. p (rolling a sum less than 6)
5. p (rolling a 7 or 11)

find the theoretical probability of each outcome:

1. p (tails) when flipping a coin once
2. p (rolling a 1) when rolling one die

Explanation:

Step1: Identify total 2-dice outcomes

Total outcomes = $6 \times 6 = 36$
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For 1. P(Rolling a sum of 10)

Step1: List favorable sum=10 pairs

Favorable pairs: $(4,6), (5,5), (6,4)$ → Count = 3

Step2: Calculate probability

$P(\text{sum}=10) = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}$
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For 2. P(Rolling doubles)

Step1: List favorable double pairs

Favorable pairs: $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$ → Count = 6

Step2: Calculate probability

$P(\text{doubles}) = \frac{6}{36} = \frac{1}{6}$
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For 3. P(Rolling a sum divisible by 3)

Step1: List sums divisible by 3

Valid sums: 3, 6, 9, 12

Step2: Count favorable pairs

Pairs for sum=3: $(1,2),(2,1)$ → 2; sum=6: $(1,5),(2,4),(3,3),(4,2),(5,1)$ →5; sum=9: $(3,6),(4,5),(5,4),(6,3)$→4; sum=12: $(6,6)$→1. Total = $2+5+4+1=12$

Step3: Calculate probability

$P(\text{sum divisible by 3}) = \frac{12}{36} = \frac{1}{3}$
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For 4. P(Rolling a sum less than 6)

Step1: List sums less than 6

Valid sums: 2, 3, 4, 5

Step2: Count favorable pairs

Pairs for sum=2: $(1,1)$→1; sum=3: $(1,2),(2,1)$→2; sum=4: $(1,3),(2,2),(3,1)$→3; sum=5: $(1,4),(2,3),(3,2),(4,1)$→4. Total = $1+2+3+4=10$

Step3: Calculate probability

$P(\text{sum}<6) = \frac{10}{36} = \frac{5}{18}$
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For 5. P(Rolling a 7 or 11)

Step1: Count favorable pairs for 7 or 11

Pairs for sum=7: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$→6; sum=11: $(5,6),(6,5)$→2. Total = $6+2=8$

Step2: Calculate probability

$P(7 \text{ or } 11) = \frac{8}{36} = \frac{2}{9}$
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For 6. P(Tails) when flipping a coin once

Step1: Identify total and favorable outcomes

Total outcomes = 2; Favorable outcomes (tails) = 1

Step2: Calculate probability

$P(\text{tails}) = \frac{1}{2}$
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For 7. P(Rolling a 1) when rolling one die

Step1: Identify total and favorable outcomes

Total outcomes = 6; Favorable outcomes (rolling 1) = 1

Step2: Calculate probability

$P(\text{rolling 1}) = \frac{1}{6}$

Answer:

1. $\frac{1}{12}$
2. $\frac{1}{6}$
3. $\frac{1}{3}$
4. $\frac{5}{18}$
5. $\frac{2}{9}$
6. $\frac{1}{2}$
7. $\frac{1}{6}$