What is the probability that a randomly selected car has a CD player?

Question

Assume that the fleet consists of 35 percent Lexus, 45 percent Infinity, and 20 percent Mercedes automobiles. 75 percent of the Lexus cars, 90 percent of the Infinities, and 40 percent of the Mercedes have CD players. Use this information to answer the following questions.
(1) What is the probability that a randomly selected car has a CD player?

(2) What is the probability that a randomly selected car either is a Mercedes or has a CD player?

Answer 1

(1) For the first problem, we need to figure out based on the information that we are given, what % of cars have CD players. To do this, we will add the parts WITH CD players as follows:

 

(.35*.75) + (.45*.9) + (.2*.4) = (.2625) + (.405) + (.08) = .7475

 

What does this tell us? That 26.25% of the total cars are Lexus with a CD player, 40.5% are Infinity with a CD player, and 8% are Mercedes with a CD player.

 

If you add up all of the cars with CD players, you get that 74.75% of the total cars, have CD players. So, you will have a 74.75% chance of selecting a car with a CD player.

 

(2) Since this is an OR probability question, and the events are not mutually exclusive, you need to use the following Prob(Event 1) + Prob(Event 2) - Prob(Event 1&2) = Prob(Event 1 OR 2), in other words:

 

          (the probability that the car is a Mercedes) + (the probability that the car has a CD player) - (the probability that the car is a Mercedes and has a CD player)

 

Has a CD player: from part (1) = 74.75%

Is a Mercedes: From the given = 20%

Is a Mercedes and has a CD player: (20%*40%)=8%

 

Total: 74.75%+20%-8%=86.75% chance that a car chosen at random with have a CD player, OR be a Mercedes.

Answer 2

Probability questions (like most word problems) tend to get us thinking of "what is the answer", and we neglect to think "how should this be organized...what approach/ruleset is needed... what is known?"

Review: 75% 0.75 3/4 they all mean that teh chance of the event occurring is three out of four, 75 of 100, on so forth.. just a ratio.

In this problem, one key is to know that probabilities for "connected events" involve multiplication. For example, the probability of tossing a fair coin twice and having both heads is 1/2 x 1/2 = 1/4. The two conncected events here are: heads, and then heads again.

Another insightful line of thought: what if the selection process is not random. What if you peek, and grab the Lexus? Of course the chances of getting the CD are 75/100 there (stated in the problem). Then, think of what the impact of not being able to peek will have.

So, random car selection: chance of it being a Lexus and it having a CD player is .35 x .75, = 0.2625. (remember- two connected events there)

Things may start to become more clear from here..
Infinity might have been selected: .45 x .90 = .405 that you get a CD
Mercedes might have been selected: .20 x .40 = .08 that you get a CD

Now just add up the individual probabilities:
.2625+.405+.08= .7475 which means almost 3/4 chance of getting a CD player if randomly selecting one of the 3 car types.

Now when I test that answer, yeah, I would expect the probability to be less than 90% since I can't always get the infinity, and I'd expect it to be way more than 40% since there are relatively few Mercedes in the pool. Notice that this problem brings to mind a "weighting" effect almost like each car type has a vote as to whether you will be listening to CDs or playing the ol' FM.