# Math Help - [SOLVED] Probability Venn Diagrams Please Double Check

1. ## [SOLVED] Probability Venn Diagrams Please Double Check

1.A survey of local car dealers revealed that 64% of all cars sold last month had CD players, 28% had alarm systems, and 22% had both CD players and alarm systems. Answer the questions below

b)What is the probability one of these cars selected at random had neither a CD player nor an alarm system? [1 point]
My answer: 1-(.42 +.22 +.06) = .70

c)What is the probability that a car had a CD player unprotected by an alarm system? [1]
My answer: .64 - .22 = .42

d)What is the probability a car with an alarm system had a CD player? [1 point]

e)Are having a CD player and an alarm system disjoint events? Are they independent? Explain. [2 points]

My thoughts:

They aren't independent because

P(both) doesn't equal P(Alarm) * P(Cd)
.22 doesnt equal .0252

They are disjoint because they are not independent and P(Alarm and Cd) does not equal 0

2. Here is the setup: $P(CD)=0.64~,~P(AS)=0.28\text{ and }P(CD~\&~AS)=0.22$.
So $P(CD\text{ or }AS)=P(CD)+P(AS)-P(CD~\&~AS)=0.70$.
That is the probability of a car having at least one of the systems.
So $1-0.70=0.30$ is the probability of a can having neither.

3. Hello, rice4lifelegit!

A survey revealed that 64% of all cars sold last month had CD players (CD),
28% had alarm systems (AS), and 22% had both CD and AS.
I assume that part (a) was to draw the Venn diagram.
Code:
      * - - - - - - - - - - - - - - - *
|                               |
|   * - - - - - - - *           |
|   | CD            |           |
|   |       * - - - + - - - *   |
|   |       |       |       |   |
|   |  42%  |  22%  |   6%  |   |
|   |       |       |       |   |
|   * - - - + - - - *       |   |
|           |            AS |   |
|   30%     * - - - - - - - *   |
|                               |
* - - - - - - - - - - - - - - - *

b )What is the probability a randomly chosen car had neither CD nor AS?
Look at your Venn diagram . . . It is 30%.

c) What is the probability that a car had a CD but not an AS?

From the diagram: . $P(CD \wedge \overline{AS}) \:=\:42\%$

d) What is the probability a car with an AS had a CD?
This one seems to be a Conditional Probability problem.

Given that the car has an AS, what is the probability that it has a CD?

$P(\text{CD } |\text{ AS}) \:=\:\frac{0.22}{0.28} \:=\:\frac{11}{14} \:\approx\:78.6\%$

d) Are having a CD and an AS disjoint events?
Are they independent? Explain.

My thoughts:

They are not independent because: . $P(\text{AS}\wedge\text{CD}) \:\neq \;P(AS)\cdot P(CD)$

They are not disjoint because: . $P(AS \wedge CD) \:\neq \:0$
Good!