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Based on a survey conducted by students in the Recreational Management degree pr...

May 14, 2024

Based on a survey conducted by students in the Recreational Management degree program, 65% of students at the university own a mountain bike, 25% own a road bike, and 6% own both a mountain bike and a road bike. a. What is the probability that a randomly selected student at this university has a mountain bike but not a road bike? b. What is the probability that a randomly selected student at this university has a road bike but not a mountain bike? c. What is the probability that a randomly selected student at this university has neither a road bike nor a mountain bike? d. Given that a randomly selected student at this university has a road bike, what is the probability he or she also has a mountain bike?

Solution by Steps

step 1

Let

P(M)

be the probability that a student owns a mountain bike,

P(R)

be the probability that a student owns a road bike, and

P(M \cap R)

be the probability that a student owns both. Given:

P(M) = 0.65

P(R) = 0.25

, and

P(M \cap R) = 0.06

step 2

The probability that a student owns a mountain bike but not a road bike is

P(M \setminus R) = P(M) - P(M \cap R)

step 3

Substituting the given values:

P(M \setminus R) = 0.65 - 0.06 = 0.59

Answer

0.59

Part (b)

step 1

The probability that a student owns a road bike but not a mountain bike is

P(R \setminus M) = P(R) - P(M \cap R)

step 2

Substituting the given values:

P(R \setminus M) = 0.25 - 0.06 = 0.19

Answer

0.19

Part (c)

step 1

The probability that a student owns neither a road bike nor a mountain bike is

P(\overline{M \cup R})

step 2

Using the principle of inclusion-exclusion:

P(M \cup R) = P(M) + P(R) - P(M \cap R)

step 3

Substituting the given values:

P(M \cup R) = 0.65 + 0.25 - 0.06 = 0.84

step 4

Therefore,

P(\overline{M \cup R}) = 1 - P(M \cup R) = 1 - 0.84 = 0.16

Answer

0.16

Part (d)

step 1

The probability that a student has a mountain bike given that they have a road bike is

P(M|R) = \frac{P(M \cap R)}{P(R)}

step 2

Substituting the given values:

P(M|R) = \frac{0.06}{0.25} = 0.24

Answer

0.24

Key Concept

Conditional Probability and Inclusion-Exclusion Principle

Explanation

The solution involves calculating probabilities using basic set operations and the principle of inclusion-exclusion. Conditional probability is used to find the likelihood of one event given another.