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Sagot :
Using the probability concept, it is found that:
a) 0.3 = 30% probability that a student at this college who takes MATH 101 will take it with Prof. B and prefer Probability.
b) 0.6 = 60% probability that a student at this college who takes MATH 101 with Prof. B will prefer Probability.
c) 0.4 = 40% probability that a student at this college who takes MATH 101 and prefers Probability is taking the course with Prof. B.
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- A probability is given by the number of desired outcomes divided by the number of total outcomes.
Item a:
- Total of 36 + 4 + 24 + 16 = 80 students.
- Of those, 24 take MATH 101 with Prof B. and prefer probability.
Thus:
[tex]p = \frac{24}{80} = 0.3[/tex]
0.3 = 30% probability that a student at this college who takes MATH 101 will take it with Prof. B and prefer Probability.
Item b:
- 24 + 16 = 40 students that MATH 101 with Prof B.
- Of those, 24 prefer probability.
Then:
[tex]p = \frac{24}{40} = 0.6[/tex]
0.6 = 60% probability that a student at this college who takes MATH 101 with Prof. B will prefer Probability.
Item c:
- 36 + 24 = 60 students prefer probability.
- Of those, 24 take the course with Prof. B.
Then:
[tex]p = \frac{24}{60} = 0.4[/tex]
0.4 = 40% probability that a student at this college who takes MATH 101 and prefers Probability is taking the course with Prof. B.
A similar problem is given at https://brainly.com/question/24658381
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