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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? [tex]P(C) = \square[/tex]
2. What is the probability of randomly choosing a Mr. Goodbar? [tex]P(G) = \square[/tex]
3. What is the probability of randomly choosing a Cookies and Cream bar? [tex]P(K) = \square[/tex]


Sagot :

Sure, let's go through the step-by-step solution for each of the probabilities from the given scenario.

### Step 1: Determine the Total Number of Candies
First, calculate the total number of miniature candy bars in the bag:

- Crunch: 15 bars
- Hershey's Chocolate: 14 bars
- Mr. Goodbar: 10 bars
- Cookies and Cream: 12 bars

Add these quantities together to get the total:

[tex]\[ 15 + 14 + 10 + 12 = 51 \][/tex]

So, there are 51 candies in total.

### Step 2: Calculate Each Probability

1. Probability of randomly choosing a Crunch bar ([tex]\(P(R)\)[/tex]):

We know there are 15 Crunch bars out of a total of 51 candies. So, the probability [tex]\(P(R)\)[/tex] is the number of Crunch bars divided by the total number of candies:

[tex]\[ P(R) = \frac{15}{51} \][/tex]

2. Probability of randomly choosing a Mr. Goodbar ([tex]\(P(G)\)[/tex]):

There are 10 Mr. Goodbars in the bag. The probability [tex]\(P(G)\)[/tex] is the number of Mr. Goodbars divided by the total number of candies:

[tex]\[ P(G) = \frac{10}{51} \][/tex]

3. Probability of randomly choosing a Cookies and Cream bar ([tex]\(P(K)\)[/tex]):

There are 12 Hershey's Cookies and Cream bars. The probability [tex]\(P(K)\)[/tex] is the number of Cookies and Cream bars divided by the total number of candies:

[tex]\[ P(K) = \frac{12}{51} \][/tex]

### Step 3: Represent Each Probability as a Reduced Fraction

- Probability of Crunch bar ([tex]\(P(R)\)[/tex]):

[tex]\[ P(R) = \frac{15}{51} \][/tex]

- Probability of Mr. Goodbar ([tex]\(P(G)\)[/tex]):

[tex]\[ P(G) = \frac{10}{51} \][/tex]

- Probability of Cookies and Cream bar ([tex]\(P(K)\)[/tex]):

[tex]\[ P(K) = \frac{12}{51} \][/tex]

So, the probabilities are as follows:

1. [tex]\(P(R) = \frac{15}{51}\)[/tex]
2. [tex]\(P(G) = \frac{10}{51}\)[/tex]
3. [tex]\(P(K) = \frac{12}{51}\)[/tex]