at sangers auto garage, three out of every five cars brought in for service need an oil change. of the cars…

at sangers 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.

at sangers 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.

Answer

Explanation:

Step1: Define probabilities

Let $P(O)$ be the probability of needing an oil - change. Given that three out of every five cars need an oil - change, so $P(O)=\frac{3}{5}$.

Step2: Define conditional probability

Let $P(T|O)$ be the probability of needing a tire - rotation given that the car needs an oil - change. Given that four out of every seven cars that need an oil - change also need a tire - rotation, so $P(T|O)=\frac{4}{7}$.

Step3: Use the formula for conditional probability

The formula for conditional probability is $P(T|O)=\frac{P(T\cap O)}{P(O)}$, and we want to find $P(T\cap O)$. Rearranging the formula gives $P(T\cap O)=P(T|O)\times P(O)$.

Step4: Calculate the probability

Substitute $P(O)=\frac{3}{5}$ and $P(T|O)=\frac{4}{7}$ into the formula: $P(T\cap O)=\frac{4}{7}\times\frac{3}{5}=\frac{12}{35}$.

Answer:

$\frac{12}{35}$