When selecting two people, find the probability that one works in an office and the other one drives a car?

Question

I have a probability question from my intro to stats class. I can't figure out if it's dependent or independent. Overall, I really don't know how to figure it out. There are 2000 people in total from a given chart; but the most important are the office workers 1050 people, and the car drivers 1150 people. I hope that's enough information. Please help!

Answer 1

This problem is equivalent to the following problem:

An urn contains 2000 balls. Each ball has two sides which each have different colors - one color on each half of the ball. 1050 of the balls have one blue side and 1150 of the balls have one red side. We don't know what the other colors are besides blue and red. What is the probability that for two balls successively selected from the urn one has a red side and the other has a blue side.

Note that we don't know how many balls have both a red and a blue side in this problem. In the absence of that information, we have to assume for purposes of computing probabilities in this problem that the colors are assigned to the balls randomly so that the probability that a ball has both a red and a blue side is the product of the two probabilities of having a blue side and a red side.


Step 1: we calculate some initial probabilities. I claim that:

P1 = P(both balls selected have one red side and one blue side) =

[(1150/2000)*(1050/2000)]*[(1149/1999)*(1049/1999)]

P2 = P(one ball has a red side but no blue and one ball has a blue side but no red) =

2*[(1150/2000)*(950/2000)]*[(1050/1999)*(850/1999)]

P3 = P(one ball has both a red and a blue side and the other has only a red side) =

2*[(1150/2000)*(1050/2000)][(1149/1999)*(950/1999)] 


P4 = P(one ball has both a red and a blue side and the other has only a blue side) =

2* [(1150/2000)*(1050/2000)][(1049/1999)*(850/1999)]


The probability that one of the balls has a red side and one has a blue side is the sum of these four probabilities.

P1 = .091054 (approximately)
P2 = .122004
P3 = .164921
P4 = .134718

So P1 + P2 + P3 + P4 = .512697 or 51.3% approximately.


So I think there is a 51.3% chance that one person has an office job and the other drives a car, given the information provided in the problem.

 

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There is another take on the problem.

 

There are between 200 and 1050 balls with both a red and a blue half.  Suppose that number is N and we know the number.

 

Then 

 

P1 = (N/2000)*((N-1)/1999)

 

P2 = 2*((1050-N)/2000)*((1150-N)/1999)

 

P3 = 2*(N/2000)((1150-N)/1999)

 

P4 = 2*(N/2000)*((1050-N)/1999)

 

So P1+P2 +P3 + P4 = (2415000 - N - N²)/(2000*1999)

 

For N = 200, the minimum, you get a probability of  59.4%

For N = 1050, the maximum, you get a probability of  32.8%

 

Notice that this argument also shows that the probability of selecting the desired combination is decreasing on the interval [200,1050].  That is, the more balls with both a red and a blue side there are the lower the probability that your selection satisfies the required conditions.

Answer 2

Since the two "events," driving and office, are independent .. one does NOT cause the other:

 

Probability of works in office = 1050/2000

Probability of driver=1150/2000

 

To put all the numbers into context (not necessary), of the 2000 people: 100 people drive AND work in office, 1/2(1050+1150-2000) ... "union" of A and B; there are 950 office workers not drivers (1050-100) ... A not B; and 1050 drivers who do not work in office ... B not A.

Answer 3

There are 2000 people.

There are 1050 office workers, so 950 drive cars.

But there are 1150 who drive cars:1150-950=200 are both office workers and drive cars.

There are 1150 car drivers, so 850 are office workers.

But there are 1050 office workers:1050-850=200 are both office workers and drive cars.

950+850=1800 people

But there are 2000 people so 200 people must be office workers and drive cars.

Looking at it three ways says there are 200 people who are office workers and who drive cars.

You have workers(W), drivers(D), and workers who drive(WD).

You choose two people.

The possibilities are W and W, W and D, D and D, W and WD, D and WD, and WD and WD.

To get a worker and a driver, you must get W and D, W and WD, D and WD, or WD and WD.

The probabilities for W and D are (1050/2000)(1150/1999)=(0.525)(0.575)=0.301875

The probabilities for W and WD are (1050/2000)(200/1999)=(0.525)(0.100)=0.052526

The probabilities for D and WD are (1150/2000)(200/1999)=(0.575)(0.100)=0.057528

The probabilities for WD and WD are (0.1)(0.09954)=0.00995

0.301875+0.052526+0.057528+0.00995=0.0.42188=42.188%  probability to choose a worker and a driver

Another thought:the probability of getting a D and W

is the same as 1-probability of not getting a D and W

The only way to do this is by getting DD or WW.

These probabilities are 0.525*0.5247=0.27546

and 0.575*0.5747=0.33035.

Add these to get 0.6059 and 1-0.6059=0.3941=39.41% which differs slightly from 42.188% !

 

 

Seeing that I posted three times and they were lost I'll try posting here.

Probability=# of favorable outcomes/total # of equally likely outcomes

Favorable outcomes=850*950=807,500

The rest of the outcomes come from WW, DD, WD WD, W WD, and D WD.

They are 360,825,  450,775,  19,900,  170,000, and 190,000.

Total of all outcomes=1,999,000

Probability=807,500/1,999,000=0.40395=40.395%