Probability A bag of miniature candy contains: - 15 Crunch bars - 14 Hershey's chocolate bars - 10 Mr. Goodbars - 12 Hershey's Cookies and Cream bars. Enter all answers as reduced fractions using / as the fraction bar. 1. What is the probability of randomly choosing a Crunch bar? $$ P(R)= $$ 2. What is the probability of randomly choosing a Mr. Goodbar? $$ P(G)= $$ 3. What is the probability of randomly choosing a Cookies and Cream bar? $$ P(K)= $$

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Let's first determine the total number of candy bars in the bag:

- **Crunch bars:** 15
- **Hershey's chocolate bars:** 14
- **Mr. Goodbars:** 10
- **Hershey's Cookies and Cream bars:** 12

**Total number of candy bars = 15 + 14 + 10 + 12 = 51**

Now, we'll calculate each probability:

1. **Probability of choosing a Crunch bar $$ P(R) $$:**
   $$
   P(R) = \frac{\text{Number of Crunch bars}}{\text{Total number of candy bars}} = \frac{15}{51}
   $$
   Simplify the fraction by dividing numerator and denominator by 3:
   $$
   P(R) = \frac{15 \div 3}{51 \div 3} = \frac{5}{17}
   $$

2. **Probability of choosing a Mr. Goodbar $$ P(G) $$:**
   $$
   P(G) = \frac{\text{Number of Mr. Goodbars}}{\text{Total number of candy bars}} = \frac{10}{51}
   $$
   The fraction $$ \frac{10}{51} $$ is already in its simplest form.

3. **Probability of choosing a Cookies and Cream bar $$ P(K) $$:**
   $$
   P(K) = \frac{\text{Number of Cookies and Cream bars}}{\text{Total number of candy bars}} = \frac{12}{51}
   $$
   Simplify the fraction by dividing numerator and denominator by 3:
   $$
   P(K) = \frac{12 \div 3}{51 \div 3} = \frac{4}{17}
   $$

**Final Answers:**
1. $$ P(R) = \frac{5}{17} $$
2. $$ P(G) = \frac{10}{51} $$
3. $$ P(K) = \frac{4}{17} $$
1. $$ P(R) = \frac{5}{17} $$ 2. $$ P(G) = \frac{10}{51} $$ 3. $$ P(K) = \frac{4}{17} $$

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