# Thread: Probability that car selected is green and has a male owner

1. ## Probability that car selected is green and has a male owner

the table shoes the results of a survey of the 120 cars in a car park, in which the colour of each car and the gender of the driver were recorded.

Green: 18 Males, 12 Females
Blue: 48 Males, 22 Females
Red: 6 Males, 14 Females

One car is selected at random.

$M$ is the event that the car selected has a male owner.
$G$ is the event that the car selected is green.
$B$ is the event that the car selected is blue.
$R$ is the event that the car selected is red.

Find $P(M\cap G)$

$P(M \cap G)=P(M)P(G)=\frac{18+48+6}{120} \times \frac{18+12}{120}= \frac{3}{20}$

2. ## Re: Probability that car selected is green and has a male owner

Originally Posted by Punch
the table shoes the results of a survey of the 120 cars in a car park, in which the colour of each car and the gender of the driver were recorded.
Green: 18 Males, 12 Females
Blue: 48 Males, 22 Females
Red: 6 Males, 14 Females
One car is selected at random.
$M$ is the event that the car selected has a male owner.
$G$ is the event that the car selected is green.
$B$ is the event that the car selected is blue.
$R$ is the event that the car selected is red.
Find $P(M\cap G)$

Look at the bit in green above.
The answer is thus $\frac{18}{120}$.

3. ## Re: Probability that car selected is green and has a male owner

Hello, Punch!

The table shows the results of a survey of the 120 cars in a car park,
in which the colour of each car and the gender of the driver were recorded.

. . $\begin{array}{c||c|c||c|} & \text{Male} & \text{Female} & \text{Total} \\ \hline \hline \text{Green} & 18 & 12 & 30 \\ \hline \text{Blue} & 48 & 22 & 70 \\ \hline \text{Red} & 6 & 14 & 20 \\ \hline \hline \text{Total} & 72 & 48 & 120 \\ \hline \end{array}$

One car is selected at random.

$M$ is the event that the car selected has a male owner.
$G$ is the event that the car selected is green.
$B$ is the event that the car selected is blue.
$R$ is the event that the car selected is red.

Find $P(M\cap G)$

$P(\text{male owner } \wedge \text{ green car}) \;=\;\frac{18}{120} \;=\;\frac{3}{20}$

(They must have used a total of 100 cars.)

4. ## Re: Probability that car selected is green and has a male owner

Thank you!

Can you help me with the next part of the question? It asks, determine whether the events M and G are independent, justifying your answer.

My answer is: Yes, since the probability of M happening or not happening is unaffected by whether G happens, M and G are independent events.

5. ## Re: Probability that car selected is green and has a male owner

Also, I suspect that the answers are all wrong...

Could you also help me check my answer for the other parts

1) $P(M \cup B)=P(M)+P(B)-P(M \cap B)=\frac{72}{120}+\frac{48+22}{120}-\frac{18+48+6}{120} \times \frac{48+22}{120}=\frac{5}{6}$

2) $P(M \cap R')= P(M)P(R')=\frac{18+48+6}{120} \times \frac{18+48+12+22}{120}=\frac{1}{2}$

But for (2) i realised that there is an alternative which uses the method used earlier on which involves just highlight the table, which gives $\frac{11}{20}$. Please advise on where I have gone wrong

6. ## Re: Probability that car selected is green and has a male owner

Your reasoning seemed so sound, it took me all day to find the flaw. Its this: The ratio of male to female car drivers is different for every car. The only reason the formulas even work for Green Cars is 18/72 == 12/48. And thus 72/120 = 18/30, and 48/120=12/48. We were mistakenly using the first numbers instead of the second. So we get a happy coincidence that makes our flawed probability generation correct. Though I use "happy" loosely since this caused me to take so long to figure it out. To see it empirically, if G=gender, C = color and we use the formula:
$P(G \cap C)= P(G)P(C)$ -- Equation 1

Using this chart:
$\begin{array}{c||c|c||c} & \text{Male} & \text{Female} & \text{Total} \\ \hline \hline \text{Green} & 18 & 12 & 30 \\ \hline \text{Blue} & 48 & 22 & 70 \\ \hline \text{Red} & 6 & 14 & 20 \\ \hline \hline \text{Total} & 72 & 48 & 120 \\ \hline \end{array}$

we get these percentages using Equation 1.

$\begin{array}{c||c|c||c} & \text{Male} & \text{Female} & \text{Total} \\ \hline \hline \text{Green} & .15 & .1 & .25 \\ \hline \text{Blue} & .35 & .2\overline3 & .58\overline3 \\ \hline \text{Red} & .1 & .0\overline6 & .1\overline6 \\ \hline \hline \text{Total} & .6 & .4 & 1.0 \\ \hline \end{array}$

Now if we multiply those percentages by 120 to get our original results back:

$\begin{array}{c||c|c||c} & \text{Male} & \text{Female} & \text{Total} \\ \hline \hline \text{Green} & 18 & 12 & 30 \\ \hline \text{Blue} & 42 & 28 & 70 \\ \hline \text{Red} & 12 & 8 & 20 \\ \hline \hline \text{Total} & 72 & 48 & 120 \\ \hline \end{array}$

Which are clearly wrong (but they do preserve the same ratio of male to female across the colors.)

The problem is we were treating the chance of being male as independent of color choice and it clearly is not. If I pick green my chance of being male is much different than if I were in a red car.

So we need 6 Percentages defined like this:
P(Male-Green) = $\frac{18}{30}$
P(Female-Green) = $\frac{12}{30}$
P(Male-Blue) = $\frac{48}{70}$
P(Female-Blue) = $\frac{22}{70}$
P(Male-Red) = $\frac{6}{20}$
P(Female-Red) = $\frac{14}{20}$

So your first question should have been:
$P(M \cap G) = P(Male-Green) P(G)$

$\frac{18}{30} \cdot \frac{30}{120} = \frac {3}{20}$

** Note $\frac{18}{30} = \frac{72}{120}$ for a "lucky" coincidence to give you the right answer from bad math.

Your final equation we use another probability identity:

$P(M \cap R') = P(M)-P(M \cap R) = P(M)-P(Red-Male)P(Red)$

$\frac{72}{120} - \frac{6}{20}\cdot \frac{20}{120} = \frac {11}{20}$

As Expected.

7. ## Re: Probability that car selected is green and has a male owner

Originally Posted by takatok
Your reasoning seemed so sound, it took me all day to find the flaw. Its this: The ratio of male to female car drivers is different for every car. The only reason the formulas even work for Green Cars is 18/72 == 12/48. And thus 72/120 = 18/30, and 48/120=12/48. We were mistakenly using the first numbers instead of the second. So we get a happy coincidence that makes our flawed probability generation correct. Though I use "happy" loosely since this caused me to take so long to figure it out. To see it empirically, if G=gender, C = color and we use the formula:
$P(G \cap C)= P(G)P(C)$ -- Equation 1

Using this chart:
$\begin{array}{c||c|c||c} & \text{Male} & \text{Female} & \text{Total} \\ \hline \hline \text{Green} & 18 & 12 & 30 \\ \hline \text{Blue} & 48 & 22 & 70 \\ \hline \text{Red} & 6 & 14 & 20 \\ \hline \hline \text{Total} & 72 & 48 & 120 \\ \hline \end{array}$

we get these percentages using Equation 1.

$\begin{array}{c||c|c||c} & \text{Male} & \text{Female} & \text{Total} \\ \hline \hline \text{Green} & .15 & .1 & .25 \\ \hline \text{Blue} & .35 & .2\overline3 & .58\overline3 \\ \hline \text{Red} & .1 & .0\overline6 & .1\overline6 \\ \hline \hline \text{Total} & .6 & .4 & 1.0 \\ \hline \end{array}$

Now if we multiply those percentages by 120 to get our original results back:

$\begin{array}{c||c|c||c} & \text{Male} & \text{Female} & \text{Total} \\ \hline \hline \text{Green} & 18 & 12 & 30 \\ \hline \text{Blue} & 42 & 28 & 70 \\ \hline \text{Red} & 12 & 8 & 20 \\ \hline \hline \text{Total} & 72 & 48 & 120 \\ \hline \end{array}$

Which are clearly wrong (but they do preserve the same ratio of male to female across the colors.)

The problem is we were treating the chance of being male as independent of color choice and it clearly is not. If I pick green my chance of being male is much different than if I were in a red car.

So we need 6 Percentages defined like this:
P(Male-Green) = $\frac{18}{30}$
P(Female-Green) = $\frac{12}{30}$
P(Male-Blue) = $\frac{48}{70}$
P(Female-Blue) = $\frac{22}{70}$
P(Male-Red) = $\frac{6}{20}$
P(Female-Red) = $\frac{14}{20}$

So your first question should have been:
$P(M \cap G) = P(Male-Green) P(G)$

$\frac{18}{30} \cdot \frac{30}{120} = \frac {3}{20}$

** Note $\frac{18}{30} = \frac{72}{120}$ for a "lucky" coincidence to give you the right answer from bad math.

Your final equation we use another probability identity:

$P(M \cap R') = P(M)-P(M \cap R) = P(M)-P(Red-Male)P(Red)$

$\frac{72}{120} - \frac{6}{20}\cdot \frac{20}{120} = \frac {11}{20}$

As Expected.
Things worked when you used, $P(M \cap G)=P(Male - Green)P(G)$

This seems to $be P(M \cap G)=P(M \cap G)P(G)$ which is not in line with any of the rules. I do not understand why the gender and colour cannot be treated as independent events when the probability that they occur or not is unaffected by the occurence or non-occurence of the other (Definition of independent event).

8. ## Re: Probability that car selected is green and has a male owner

Because probability is a tricksy hobbit. The probability of being male or buying a green car is in fact NOT independent in this case. You are thinking (like I did) of random people walking in being male or female.. then randomly buyng a car. If I stated the problem as this: The odds of someone walking in and buying one of these 3 colors is this: P(g), P(r), P(b). Then i said the odds of a person's gender are this: P(m), P(f). We would indeed have 2 independent events, and we could calculate all the odds like you said. In fact given those probabilities up front, we could say how far off are actual data is from the expected return. And in fact the expected ratio of male to female buyers in each color should be EXACTLY the same in that case. Specifically the ratio of P(m) to P(f).

However, we are given ZERO probabilities. Instead we are given a set of data, and asked to fill in probabilities from that. Since the ratio of male to female is in no way the same for any of the 3 data points (red,blue and green). We have to make up probabilities for each. So in this case since we are working backwards from established data, the 2 events are very dependent on each other. If they are buying a green car P(m) and P(f) are very different than if they were buying a red car. The very definition of dependent. This problem was odd because it was "backward" from most problems. Instead of finding the odds of Events happening, we instead got a set of data points that said these events DID happen. Please find some P(x) that makes this data point the expected value.

Another confusing thing is you (and Me too trying to fit in with what you were doing) was trying to use the actual probability rules. Which work as long as we agree that what I said above is true and used my modified %. Plato cut right to the chase and realized immediately all this is doing is asking you to read a chart to find the P(x) to that makes the expected value in the chart. For example the $P(M \cap R')$, We should just read the chart and say 18+48/120 = 11/20. No need to go the long way around

Hopefully I am explaining it clearly enough that you can follow that. If not let me know and I'll try to come up with another line of attack.

9. ## Re: Probability that car selected is green and has a male owner

Originally Posted by takatok
Because probability is a tricksy hobbit. The probability of being male or buying a green car is in fact NOT independent in this case. You are thinking (like I did) of random people walking in being male or female.. then randomly buyng a car. If I stated the problem as this: The odds of someone walking in and buying one of these 3 colors is this: P(g), P(r), P(b). Then i said the odds of a person's gender are this: P(m), P(f). We would indeed have 2 independent events, and we could calculate all the odds like you said. In fact given those probabilities up front, we could say how far off are actual data is from the expected return. And in fact the expected ratio of male to female buyers in each color should be EXACTLY the same in that case. Specifically the ratio of P(m) to P(f).

However, we are given ZERO probabilities. Instead we are given a set of data, and asked to fill in probabilities from that. Since the ratio of male to female is in no way the same for any of the 3 data points (red,blue and green). We have to make up probabilities for each. So in this case since we are working backwards from established data, the 2 events are very dependent on each other. If they are buying a green car P(m) and P(f) are very different than if they were buying a red car. The very definition of dependent. This problem was odd because it was "backward" from most problems. Instead of finding the odds of Events happening, we instead got a set of data points that said these events DID happen. Please find some P(x) that makes this data point the expected value.

Another confusing thing is you (and Me too trying to fit in with what you were doing) was trying to use the actual probability rules. Which work as long as we agree that what I said above is true and used my modified %. Plato cut right to the chase and realized immediately all this is doing is asking you to read a chart to find the P(x) to that makes the expected value in the chart. For example the $P(M \cap R')$, We should just read the chart and say 18+48/120 = 11/20. No need to go the long way around

Hopefully I am explaining it clearly enough that you can follow that. If not let me know and I'll try to come up with another line of attack.
I agree that the aim of the question is to read a chart to find the P(x). Interesting to see your workings and thoughts. A very big thank you to U!