conditional probability problem

[lila]

Any thoughts on this one?

The following data from a sample of 100 show the record of car ownership of fathers and sons. In 22 families, both father and son own a vehicle; in 31 families, neither father nor son own vehicles; in 12 families, the father owns a vehicle while the son does not; and in 35 families, the son owns a car while the father does not.

what is the probability that a son owns a car given that his father owns a car?
what is the probability a son owns a car given his father does not own a car?

The first thing I tried to do is come up with probabilities for each event. For the probability of the father owning a car, I came up with P (F) = 12/100. For the probability of the son owning a car, I came up with P (S) = 35/100.

When I tried it, I first tried to find P(F U S), thinking that would lead me to the answer to the first question. I came up with 0.25, but that isn't the right answer.

When I tried the second half of the question, I tried P (F U S) - P (F), and came up with 0.35, but that isn't the right answer either. What am I doing wrong?

 

[tkhunny]

The following data from a sample of 100 show the record of car ownership of fathers and sons. In 22 families, both father and son own a vehicle; in 31 families, neither father nor son own vehicles; in 12 families, the father owns a vehicle while the son does not; and in 35 families, the son owns a car while the father does not.

Just draw a diagram. A big box with two intersecting circles is all it takes.

The first thing I tried to do is come up with probabilities for each event. For the probability of the father owning a car, I came up with P (F) = 12/100. For the probability of the son owning a car, I came up with P (S) = 35/100.

12/100 is no good. You have not included the other 22. 35/100 has the same problem. Those 22 still are missing.

what is the probability that a son owns a car given that his father owns a car?

Father owns a car: 12 + 22 = 34 Son ALSO owns a car: From those 34, there are 22 Thus 22/34 is the conditional probability.

what is the probability a son owns a car given his father does not own a car?

Father has no car: 31 + 35 = 66 Son DOES own a car: From those 66, there are 35. Thus, 35/66 is the conditional probability.

 

[arthur ohlsten]

let F be the father owns a car
let S be that the son owns a car
let P be the population of 100
_ _
P=[F + F] [S + S]
_ _ __
P=FS +FS + FS + FS
_ _ __
FS=22.....FS=12.....FS=35 .....FS=31

a) probability son has a car knowing father has a car is:
_
FS/[FS+FS] = 22/[22+12]
_ _ __
FS / [FS+FS = 35/[12+31]

please check for errors

Arthur

 

[galactus]

If you make a chart, it can help immensely.

+-----+--------+---------+-------+
|     | Father | Father  |       |
|     |  own   | not own | total |
+-----+--------+---------+-------+
| son |        |         |       |
| own |   22   |   35    |   57  |
+-----+--------+---------+-------+
| son |        |         |       |
| not |   12   |   31    |   43  |
| own |        |         |       |
+-----+--------+---------+-------+
| tot |   34   |   66    |  100  |
+-----+--------+---------+-------+

For instance, probability son doesn't own given father doesn't own.

Use the son doesn't own and go down the 'given' column or row. 31/66

Probability father owns given son doesn't own. 12/43

 

[lila]

Thank you, that makes it much clearer to me. This stuff is actually pretty fun to think about, once you take the time to figure out the logic. Thanks again!

 

[galactus]

I am assuming Soroban fixed me up. Thanks buddy.