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At Sanger’s auto garage, three out of every five cars brought in for service need an oil change. Of the cars that need an oil change, four out of every seven also need a tire rotation. What is the probability that a car that comes into the garage needs both an oil change and a tire rotation? Give the answer in fraction form.

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To determine the probability that the car from Sanger's auto garage needs an oil change and also tire rotation, we multiply the probability of the first event and the second event. We get an equation of,
                                     = (3/5) x (4/7) = 12/35
Thus, the probability that the car needs both oil change and tire rotation is 12/35. 

Using conditional probability, it is found that there is a [tex]\frac{12}{35}[/tex] probability that a car that comes into the garage needs both an oil change and a tire rotation.

What is Conditional Probability?

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem, the events are:

  • Event A: Car needs an oil change.
  • Event B: Car needs a tire rotation.

The probabilities are:

  • Three out of every five cars brought in for service need an oil change, hence [tex]P(A) = \frac{3}{5}[/tex].
  • Four out of every seven also need a tire rotation, hence [tex]P(B|A) = \frac{4}{7}[/tex].

Then:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(A)P(B|A)[/tex]

[tex]P(A \cap B) = \frac{3}{5} \times \frac{4}{7} = \frac{12}{35}[/tex]

[tex]\frac{12}{35}[/tex] probability that a car that comes into the garage needs both an oil change and a tire rotation.

You can learn more about conditional probability at https://brainly.com/question/14398287

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