Math  /  Data & Statistics

QuestionLouisa Pectol 11/25/24 9:20 PM rule and trees Question 7, 14.CE. 24 HW Score: 60%,660 \%, 6 of 10 points Points: 0 of 1
Twenty percent of cars that are inspected have faulty pollution control systems. The cost of repairing a pollution control system exceeds $150\$ 150 about 10%10 \% of the time. When a takes her car in for inspection, what is the probability that she will end up paying more than $150\$ 150 to repair the pollution control system?

Studdy Solution

STEP 1

What is this asking? What's the chance Louisa's car both fails inspection *and* needs expensive repairs? Watch out! Don't just add the percentages, we're looking for the probability of *both* events happening!

STEP 2

1. Define Events and Probabilities
2. Calculate Joint Probability

STEP 3

Let's **define** what we know!
We'll call failing the inspection event FF, so P(F)=0.20P(F) = 0.20 (that's 20100\frac{20}{100}, the **20 percent** given).

STEP 4

Let's call needing expensive repairs (over $150\$150) event EE.
We're told P(EF)=0.10P(E|F) = 0.10 (**10 percent**).
This is the probability of expensive repairs *given* that the car failed inspection.

STEP 5

We want to find P(EF)P(E \cap F), the probability that *both* EE and FF occur.
Remember the formula for **conditional probability**: P(EF)=P(EF)P(F) P(E|F) = \frac{P(E \cap F)}{P(F)}

STEP 6

We can **rearrange** this formula to solve for what we want: P(EF)=P(EF)P(F) P(E \cap F) = P(E|F) \cdot P(F) This makes sense, right?
The chance of *both* happening is the chance of one happening *times* the chance of the other happening *given* the first one happened.

STEP 7

Now, we just **plug in** our **values**: P(EF)=(0.10)(0.20) P(E \cap F) = (0.10) \cdot (0.20)

STEP 8

**Multiply** those probabilities! P(EF)=0.02 P(E \cap F) = 0.02

STEP 9

Louisa has a 0.02\textbf{0.02} (or **2%**) chance of needing to pay more than $150\$150 to fix her pollution control system after the inspection.

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