Hi,
I have got following question:

A jar has 5 marbles, 1 of each of the colors red, white, blue, green and yellow. If 4 marbles are removed from the jar, what is the probability (i) that the yellow one was removed? (ii) that the yellow one was not removed?


probability = favorable outcomes/total outcomes

I think the total outcomes is 4 because at a time we are picking 4 marbles.Now the probability that it is removed is 1/4, because out of 4 only 1 would be the yellow.
And the probability that it is not removed is: 4/4 because it is possible that out of 4 none of them is yellow.

Answer is not correct so please guide me.

Zulfi.

 

And the probability that it is not removed is: 4/4 because it is possible that out of 4 none of them is yellow..

Let's suppose you've correctly figured out (i). What do you know about the sum of (i) an (ii)?

As for the total outcomes, you might consider trying this in real life. I think you'll soon discover there are many more ways to draw 4 marbles from a pool of 5.

Using the language of statistics, what category of problem is this? Have you studied permutations for instance?

 

Probabilities can be difficult to solve until you learn how to think about the problem. This is why I asked what your answer is to the earlier lunch problem . I don't think that you yet understand that problem and when you move forward, it gets more difficult if you don't understand the earlier material.

For any event, the probability of its occurrence plus the probability of its nonoccurence is exactly 1. This is true all the time for every event.

For example, flip a coin - probability of it coming up heads=.5 and the probability of it NOT coming up heads=.5 and, like I said, the two probabilities add up to exactly 1.

If you can answer these questions without overthinking them, you will be much closer to understanding how to answer the homework question:

What is the probability of picking the yellow marble, if you only pick one marble out of the 5 that are in the jar? What is the probability of NOT picking the yellow marble, if you only pick one marble out of the 5 that are in the jar?

Make two columns, put the probability of picking yellow in one column and the probability of not picking yellow in a second column.

Now, answer the question: What is the probability of picking the yellow marble, if you only pick one marble out of 4 that are in the jar? What is the probability of NOT picking the yellow marble, if you only pick one marble out of 4 that are in the jar? Write down the probabilities.

What is the probability of picking the yellow marble, if you only pick one marble out of 3 that are in the jar? What is the probability of NOT picking the yellow marble, if you only pick one marble out of 3 that are in the jar? Write down the probabilities.

What is the probability of picking the yellow marble, if you only pick one marble out of 2 that are in the jar? What is the probability of NOT picking the yellow marble, if you only pick one marble out of 2 that are in the jar? Write down the probabilities.

For the moment, don't get hung up on thinking about what happens if you pick the yellow one first. Just assume for the moment that for each of the above events, the yellow marble is in the jar when you pick a marble.

What are your answers?

 

Hi,

When the problem is simple enough you can just sit down and start reeling out the actual possibilities one by one, then observe what is happening in a general way.

For example, say you have only three marbles numbered 1,2, and 3.
You can make picks as follows:
1 and 2
1 and 3
2 and 1
2 and 3
3 and 1
3 and 2

but since some of these are just repeats in a different order, we can truncate the list down to:
1 and 2
1 and 3
2 and 3

Expanding to show the one that remains in the jar [n] after a pick, we have:
1 and 2, [3]
1 and 3, [2]
2 and 3, [1]

See if you can figure out the probability that you will pick say #2 and the probability that #2 will remain in the jar.
This is the same problem but simpler because there are less possibilities, but when there are not too many possibilities (as with 5 marbles) you can think in the same manner and make a list like that above.

 

Hi,
I have got following question:

A jar has 5 marbles, 1 of each of the colors red, white, blue, green and yellow. If 4 marbles are removed from the jar, what is the probability (i) that the yellow one was removed? (ii) that the yellow one was not removed?


probability = favorable outcomes/total outcomes

I think the total outcomes is 4 because at a time we are picking 4 marbles.Now the probability that it is removed is 1/4, because out of 4 only 1 would be the yellow.
And the probability that it is not removed is: 4/4 because it is possible that out of 4 none of them is yellow.

Answer is not correct so please guide me.

Zulfi.

If the probability that it is not removed is 4/4, that means that it can't be removed.

The yellow marble is either removed or it is not removed. Those are the only two possibilities, right? It can't be both removed and not removed. It can't be neither removed nor not removed. It must be exactly one or the other. So what do you know about sum of the probability that it is removed and the possibility that it isn't?

Think about the chain of events that leads to the yellow marble not being removed.

You remove the first marble. What are the chances that you did NOT remove the yellow marble?
You remove the second marble. What are the chances that you did NOT remove the yellow marble?
You remove the third marble. What are the chances that you did NOT remove the yellow marble?
You remove the fourth marble. What are the chances that you did NOT remove the yellow marble?

What is the probability of all four of these things happening one after another?

 

Hi,

When the problem is simple enough you can just sit down and start reeling out the actual possibilities one by one, then observe what is happening in a general way.

For example, say you have only three marbles numbered 1,2, and 3.
You can make picks as follows:
1 and 2
1 and 3
2 and 1
2 and 3
3 and 1
3 and 2.

You ever play exacta boxes

 

You ever play exacta boxes

Hi,

No, have you?
What is the significance of that?

 

Hi,

No, have you?
What is the significance of that?

It is a kind of bet in horse racing and you explained it rather well.

 

Hi,
I have got following question:

A jar has 5 marbles, 1 of each of the colors red, white, blue, green and yellow. If 4 marbles are removed from the jar, what is the probability (i) that the yellow one was removed? (ii) that the yellow one was not removed?


probability = favorable outcomes/total outcomes

I think the total outcomes is 4 because at a time we are picking 4 marbles.Now the probability that it is removed is 1/4, because out of 4 only 1 would be the yellow.
And the probability that it is not removed is: 4/4 because it is possible that out of 4 none of them is yellow.

Answer is not correct so please guide me.

Zulfi.

In this particular case, you could always solve (ii) first.
Total outcomes = total ways to pick 4 marbles ( Try to list out all the cases or think combinatrics)
Favourable outcomes ( yellow is not picked) = 1 ( try to think of all the cases which meet this criteria)

Now you can simply apply the probability formula.
For i) take a look at Raymonds comment.

Hope I helped

 

Hi everybody,

I got some understandability.

For Mr. Raymonds:

What is the probability of picking the yellow marble, if you only pick one marble out of the 5 that are in the jar?

1/5, because there is only one yellow marble.

What is the probability of NOT picking the yellow marble, if you only pick one marble out of the 5 that are in the jar?

4/5 because there are 4 non yellow marbles.


What is the probability of picking the yellow marble, if you only pick one marble out of 4 that are in the jar?

Sir, you have changed the question. I don’t know how many yellow marbles are in jar. If there is only one yellow then it would ¼

What is the probability of NOT picking the yellow marble, if you only pick one marble out of 3 that are in the jar?

Again I am supposing only one yellow marble in the jar. Then it would be: 2/3

What is the probability of picking the yellow marble, if you only pick one marble out of 2 that are in the jar?

½

What is the probability of NOT picking the yellow marble, if you only pick one marble out of 2 that are in the jar?

½.
<
Expanding to show the one that remains in the jar [n] after a pick, we have:
1 and 2, [3]
1 and 3, [2]
2 and 3, [1]

See if you can figure out the
>
probability that you will pick say #2 = 1/3 because there is only one #2
and the probability that #2 will remain in the jar.
(i would tell you if i #2 is not picked = 2/3 because there are 2 other marbles)

Thanks all.
God bless you all for helping me and others.
Zulfi.

 

Again, think of the chain of events needed to draw four marbles out of the back and end up with a yellow marble in the bag.

Step 1:

You have a bag with five marbles, exactly one of which is yellow.
What is the probability of drawing one marble from the bag and having it NOT be yellow.
If this happens, then you are still on the path to the outcome being sought -- so continue.
How many marbles are now in the bag?
How many of them are yellow?
What is the overall probability of getting to this point?

Step 2:

Given the bag as it ended up from the previous step.
What is the probability of drawing one marble from the bag and having it NOT be yellow.
If this happens, then you are still on the path to the outcome being sought -- so continue.
How many marbles are now in the bag?
How many of them are yellow?
What is the overall probability of getting to this point?

Step 3:

Given the bag as it ended up from the previous step.
What is the probability of drawing one marble from the bag and having it NOT be yellow.
If this happens, then you are still on the path to the outcome being sought -- so continue.
How many marbles are now in the bag?
How many of them are yellow?
What is the overall probability of getting to this point?

Step 4:

Given the bag as it ended up from the previous step.
What is the probability of drawing one marble from the bag and having it NOT be yellow.
If this happens, then you are still on the path to the outcome being sought -- so continue.
How many marbles are now in the bag?
How many of them are yellow?
What is the overall probability of getting to this point?

How does the bag no compare with the outcome being sought?

 

Hi everybody,

I got some understandability.

For Mr. Raymonds:

What is the probability of picking the yellow marble, if you only pick one marble out of the 5 that are in the jar?

1/5, because there is only one yellow marble.

What is the probability of NOT picking the yellow marble, if you only pick one marble out of the 5 that are in the jar?

4/5 because there are 4 non yellow marbles.


What is the probability of picking the yellow marble, if you only pick one marble out of 4 that are in the jar?

Sir, you have changed the question. I don’t know how many yellow marbles are in jar. If there is only one yellow then it would ¼

What is the probability of NOT picking the yellow marble, if you only pick one marble out of 3 that are in the jar?

Again I am supposing only one yellow marble in the jar. Then it would be: 2/3

What is the probability of picking the yellow marble, if you only pick one marble out of 2 that are in the jar?

½

What is the probability of NOT picking the yellow marble, if you only pick one marble out of 2 that are in the jar?

½.
<
Expanding to show the one that remains in the jar [n] after a pick, we have:
1 and 2, [3]
1 and 3, [2]
2 and 3, [1]

See if you can figure out the
>
probability that you will pick say #2 = 1/3 because there is only one #2
and the probability that #2 will remain in the jar.
(i would tell you if i #2 is not picked = 2/3 because there are 2 other marbles)

Thanks all.
God bless you all for helping me and others.
Zulfi.

Good deal, you are starting to think about it and you are bringing up questions that show you are thinking about it.

For simplicity, rephrase your homework question from “A jar has 5 marbles, 1 of each of the colors red, white, blue, green and yellow.” To, “A jar has 5 marbles, 1 yellow marble and 4 non-yellow marbles.” You can do this because you only care whether a marble is yellow or not. We will call the marble either Y for yellow or NY for not yellow.

Here is something that you have to accept: When your problem says “4 marbles are removed”, it means, for our purposes, removing the marbles one at a time. Note the important point that you are removing the marbles without ever putting them back. This is called “without replacement”.

You want to know - If 4 marbles are removed from the jar, what is the probability (i) that the yellow one was removed? (ii) that the yellow one was not removed?

So, some of your questions are going to be along the lines of, well what happens if I picked the yellow marble first – why am I still picking? The answer has to do with a conditional probability and accounting for all possible outcomes.

To solve this you need to think of it as a set of paths of picking 4 marbles one at a time and account for all of the possibilities.

Let me start you off:

Y 1/5 (.20)

Pick 1 –

NY 4/5 (.80)

As you already answered, on your first pick, the probability of getting a yellow is 1/5 (.20) and not getting a yellow is 4/5 (.80).

Note that .20+.80 = 1 because, as we said, the probability of any event occurring + the probability of it not occurring has to equal 1.

Now on to Pick 2

But now we have to start drawing a tree, because we have to have different paths representing the two possible outcomes for pick 1. Except, there is only one path when you already got yellow on the first pick - there is only the possibility of getting NY and that probability = 1 because it can only be NY.

Pick 2 – when P1=Y (.20), P2=NY 4/4 (1.0)

When you did not pick the yellow marble in pick 1, you have two possible outcomes for pick 2.

Pick 2 - when P1=NY(.80)

P2=Y 1/4 (.25)
P2=NY 3/4 (.75)

Now, I have accounted for all possible outcomes of picking two marbles and I can answer the question of the probability of getting a yellow in two picks and of not getting a yellow in two picks.

Here are your paths:

1. P1=Y, P2=NY

2. P1=NY, P2=Y

3. P1=NY, P2=NY

For each path, you multiply the probabilities for each branch to get the probability of that path:

1. P1=Y(.20), P2=NY(1.0) – probability of this this path = .20 x 1.0 = .20

2. P1=NY (.80), P2=Y (.25) – probability of this path = .80 x .25 = .20

3. P1=NY (.80), P2=NY (.75) – probability of this path = .80 x .75 = .60

You can see that you have accounted for all paths because the probabilities add up to 1.0 and that is an important check that you are doing it correctly.

Two paths result in getting a yellow marble and, therefore the probability of getting a yellow marble, in two picks = .40 (.20+.20). Probability of not getting a yellow marble in two draws=.60.

You can now start a gambling operation where you will pay someone twice their bet if they draw a yellow marble and you will give them two draws to get it!. You will win 60% of the time and make a fortune!

OK, I’m kidding, but what you can do is continue the tree for three picks and then four picks. Work slowly and surely, draw out the branches, and get the probability of each path. Add the probabilities for paths that contain the yellow marble and those that do not contain the yellow marble. Check that, together, they equal 1. Now you have the answers……which I am eagerly waiting for


small edit for clarity (how do you get strikeout?)

 

Hi,
WBhan thanks for your set of questions. But i cant reply you. I already answered questions of Mr Raymond & MrAI. My learning is more formal. I am reading from the book and if get stuck, i try to seek knowledge from you people. This is how i am learning under your and under the guidance of other learned people in this forum. I would try to solve questions posed by you people but not often. This is because i wont concentrate on the book. I would come back to your questions when i would get time.

Zulfi.

 

Hi,
WBhan thanks for your set of questions. But i cant reply you. I already answered questions of Mr Raymond & MrAI. My learning is more formal. I am reading from the book and if get stuck, i try to seek knowledge from you people. This is how i am learning under your and under the guidance of other learned people in this forum. I would try to solve questions posed by you people but not often. This is because i wont concentrate on the book. I would come back to your questions when i would get time.

Zulfi.

That's fine. I provided you with a formal chain of reasoning that leads directly to the solution to the problem you asked about. If you don't think it is worth your time to consider, that's completely your choice. Good luck to you.

 

Hi everybody,

I got some understandability.


START QUOTE of MRAL:
Expanding to show the one that remains in the jar [n] after a pick, we have:
1 and 2, [3]
1 and 3, [2]
2 and 3, [1]

See if you can figure out the probability that you will pick say #2 = 1/3 because there is only one #2
and the probability that #2 will remain in the jar.

END QUOTE of MRAL

(i would tell you if i #2 is not picked = 2/3 because there are 2 other marbles)

Thanks all.
God bless you all for helping me and others.
Zulfi.

Hello again,

The three distinct cases were:
1 and 2, [3]
1 and 3, [2]
2 and 3, [1]

These represent three different experiments. If we do all three, we picked 1 twice, we picked 2 twice, and we picked 3 twice. Also, at the end of the three experiments we always had one marble left and it was either 1, 2, or 3.
Thus for the three experiments we always picked the same marble twice, and we left a single marble in the jar but it was never the same marble.
So after the three experiments we picked #2 twice out of 3 experiments, and we left #2 once out of three experiments. So the ratios are 2/3 and 1/3 respectively.
Now try to apply that to the 5 marble case.

As i said before this kind of method works very well when there are not that many possibilities and i think it satisfies the intuition. If we had 100 marbles in the jar we would have to look to theory to get the right result or we'd have to make a very big list.

 

Hi,
/--. I already answered questions of Mr Raymond & MrAI. My learning is more formal. I am reading from the book and if get stuck, i try to seek knowledge from you people. -//
Zulfi.

Hi @zulfi100,

I think that you have been respectful and appreciative regarding your requests for guidance.

Speaking only for myself, I have asked you questions to get a sense of where you are with your understanding and lack of understanding that led to the wrong answer and your request for guidance. My intention is that if you answer the questions I ask, it might in turn guide your thinking to see the solution...that is the hope, but I know it doesn't always work that way.

Again, speaking only for myself, I would appreciate the feedback of knowing where you are with solving the problem. Did you now get the correct answer for the right reason? Did you give up and move on to the next question?