Order matter or not (probability question)

[evv89]

Hello everyone,
I really got stuck in slot method application. Please, help me to clarify order matters or not step.

In one of the CAT problem order matters:

The Carson family will purchase three used cars. There are two models of cars available, Model A and Model B, each of which is available in four colors: blue, black, red, and green. How many different combinations of three cars can the Carsons select if all the cars are to be different colors?

So here order does not matter and we eventually divide by 3!


BUT in another CAT problem I see the contrary, please see my comments for solution explanation in red :

A jar contains 8 red marbles and y white marbles. If Joan takes 2 random marbles from the jar, is it more likely that she will have 2 red marbles than that she will have one marble of each color?

(1) y ≤ 8
(2) y ≥ 4

In total, there are 8 red marbles, y white marbles, and 8 + y total marbles in the jar. The probability of obtaining two red marbles is given by:

P(Red AND Red) =

8

8 + y

×

7

7 + y

=

56 So it means that order matter if we do not divide by 2???? But what is the difference in order if we need to pick 2 reds marbles (identical)???

(8 + y) × (7 + y)

The probability of obtaining one red marble and one white marble is given by:

P(Red AND White) =
8

8 + y

×
y

7 + y

× 2 =
16y In this problem we also use slot method, but multiple by 2 to make the situation in which order matters. Is it because we construct this slot from two subsets (theredore the order is not counted???? Please, confirm.

(8 + y) × (7 + y)

[tim]

Okay, if i'm interpreting things correctly, it looks like your question boils down to why we have to account for order when drawing a red and a white marble but not when drawing two reds. Ultimately it comes to this:

If you are drawing one of each, you will either need to get a red and then a white OR a white and then a red. You can calculate these two probabilities separately and then add them (because of the OR), or if you recognize that the probabilities are the same you can just calculate one of them and then multiply by 2..

If you are drawing two red marbles, there is only one way to do that - a red and then a red. No need to multiply by anything here..

[phmphuoc]

Okay, if i'm interpreting things correctly, it looks like your question boils down to why we have to account for order when drawing a red and a white marble but not when drawing two reds. Ultimately it comes to this:

If you are drawing one of each, you will either need to get a red and then a white OR a white and then a red. You can calculate these two probabilities separately and then add them (because of the OR), or if you recognize that the probabilities are the same you can just calculate one of them and then multiply by 2..

If you are drawing two red marbles, there is only one way to do that - a red and then a red. No need to multiply by anything here..

Hi Tim,
To my understanding, What evv89 want to ask you that why drawing 2 reds we do not divide its probability by 2 because the order does not matter in this case!

I have the same concern as his. I think the probability of drawing 2 reds out of 8 should be 8C2 = 28.

Please correct me!
Thanks

[tim]

if i'm interpreting your question correctly, you're asking why we don't divide by 2 when there are two reds. i've already explained why we have to multiply by 2 when you have one of each of two colors; this accounts for the difference between the answers for two of the same color and the answers for two of different colors. dividing by 2 as you suggest would effectively account for the difference twice..

[pranabiitkgp]

Let me try to ans this .
Here we can construct 3 cases out of these 2 qs.
If we carefully look at the final qs asked then we will see :
1> finally we are interested to find out certain outcome where there are cars with 3 different colors . How those things appear not inportant and does not change the no. of ways those set of 3 different cars can be chosen .

2>How many ways 2 marbels can be selected . In what order we are selecting the marbels will not change the outcome.

3> One each of different colors . Here comes the important of the order . Suppose there are 3 red and 3 blue marbel - if we select red first then there will be one fewer marbel in the lot hence probality of getting blue marbel will be different if we select red first . and vis a vis . As we dont know whether we r selecting red first or blue first we are multiplying by 2 at the end .

hope this will clarify .

Thanks,
PM.

[tim]

let us know if there are any further questions on this one..

[ab.fakhry]

I really don't understand why!!

I believe it's simple as this:

it is more likely that she will have 2 red marbles than that she will have one marble of each color only IF the white marbles are LESS than 8.

Is this logical??

[tim]

no, this is not logical. you have not connected your reasoning to the conclusion at all..

[jmacavoy]

Okay, if i'm interpreting things correctly, it looks like your question boils down to why we have to account for order when drawing a red and a white marble but not when drawing two reds. Ultimately it comes to this:

If you are drawing one of each, you will either need to get a red and then a white OR a white and then a red. You can calculate these two probabilities separately and then add them (because of the OR), or if you recognize that the probabilities are the same you can just calculate one of them and then multiply by 2..

If you are drawing two red marbles, there is only one way to do that - a red and then a red. No need to multiply by anything here..

If two (or more) events are mutually exclusive, we can find the probability that one or the other of them happens by adding the probabilities of each of them. Here is a helpful refresher from a UC Berkeley stats class: http://www.google.com/url?q=http://www.stat.berkeley.edu/~bradluen/stat2/lecture16.pdf&sa=U&ei=hWsSUdyuKJT7yAH9r4HQDA&ved=0CB0QFjAA&usg=AFQjCNE8_u8I5w0Ck5gJHDBAcYjUVJsiCQ

[tim]

cool