In a freshman high school class of 80 students, 22 students take Consumer Education, 20 students take French, and 4 students take both. Which equation can be used to find the probability, [tex]\(P\)[/tex], that a randomly selected student from this class takes Consumer Education, French, or both?

A. [tex]\(P = \frac{11}{40} + \frac{1}{4} - \frac{1}{20}\)[/tex]

B. [tex]\(P = \frac{11}{40} + \frac{1}{4} - \frac{1}{10}\)[/tex]

C. [tex]\(P = \frac{11}{40} + \frac{1}{4}\)[/tex]

D. [tex]\(P = \frac{11}{40} + \frac{1}{4} + \frac{1}{20}\)[/tex]



Answer :

To find the probability that a randomly selected student from the class takes Consumer Education, French, or both, we need to use the principle of inclusion-exclusion. Here's the step-by-step solution:

1. Determine the total number of students:
Total students = 80

2. Calculate individual probabilities:
- Number of students taking Consumer Education = 22
[tex]\( P(\text{Consumer Education}) = \frac{22}{80} \)[/tex]
- Number of students taking French = 20
[tex]\( P(\text{French}) = \frac{20}{80} \)[/tex]
- Number of students taking both Consumer Education and French = 4
[tex]\( P(\text{Both}) = \frac{4}{80} \)[/tex]

3. Convert these probabilities to decimals (this step is only for clarity, we can use fractions directly in calculations):
[tex]\( P(\text{Consumer Education}) = \frac{22}{80} = 0.275 \)[/tex]
[tex]\( P(\text{French}) = \frac{20}{80} = 0.25 \)[/tex]
[tex]\( P(\text{Both}) = \frac{4}{80} = 0.05 \)[/tex]

4. Apply the inclusion-exclusion principle:
The formula to find the probability of either event (taking Consumer Education, taking French, or taking both) is:
[tex]\[ P(\text{Consumer Education or French}) = P(\text{Consumer Education}) + P(\text{French}) - P(\text{Both}) \][/tex]
Plugging in the numbers:
[tex]\[ P(\text{Consumer Education or French}) = 0.275 + 0.25 - 0.05 = 0.475 \][/tex]

5. Convert the result to a fraction (for comparison with given options):
To match the given options, let's express each probability with a common denominator. First, note that the common denominator for the fractions provided is 40.

- [tex]\( P(\text{Consumer Education}) = \frac{22}{80} = \frac{11}{40} \)[/tex]
- [tex]\( P(\text{French}) = \frac{20}{80} = \frac{1}{4} = \frac{10}{40} \)[/tex]
- [tex]\( P(\text{Both}) = \frac{4}{80} = \frac{1}{20} = \frac{2}{40} \)[/tex]

Now, use these fractions in the inclusion-exclusion formula:
[tex]\[ P = \frac{11}{40} + \frac{10}{40} - \frac{2}{40} = \frac{11 + 10 - 2}{40} = \frac{19}{40} = 0.475 \][/tex]

6. Compare with given options:
- [tex]\( P = \frac{11}{40} + \frac{1}{4} - \frac{1}{20} = 0.475 \)[/tex] (matches our 0.475)
- [tex]\( P = \frac{11}{40} + \frac{1}{4} - \frac{1}{10} \approx 0.425 \)[/tex]
- [tex]\( P = \frac{11}{40} + \frac{1}{4} \approx 0.525 \)[/tex]
- [tex]\( P = \frac{11}{40} + \frac{1}{4} + \frac{1}{20} \approx 0.575 \)[/tex]

Therefore, the correct equation that can be used to find the probability [tex]\( P \)[/tex] is:
[tex]\[ P = \frac{11}{40} + \frac{1}{4} - \frac{1}{20} \][/tex]