Choosing multiple probabilities

[JeffM]

Let's look at your first answer of 1320. I do not agree. Let's label the pears a, b, c, d, e, f, g, h, i, j, k, m, and n.

If the order in which you draw your sample matters, then you are correct that the number of possible samples is

[MATH]\dfrac{12!}{(12 - 3)!}[/MATH].

But in fact you do not care what order you select a, b, and c. That represents

[MATH]3![/MATH] of each ordered sample.

So your first answer should be [MATH]\dfrac{12!}{(12 - 3)! * 3!} = \dfrac{1320}{6} = 220.[/MATH].

 

[pka]

"(a) In how many ways can a sample of 3 pears be taken from a basket containing 12 pears? Part 2: If 2 pears are spoiled, in how many ways can the sample of 3 include (b) 1 spoiled pear (c) no spoiled pears (d) at most 1 spoiled pear."

Suppose that \(X\) is the number of spoiled pears in a selection of three made from a basket containing twelve, two of which are soiled.
Now, the possible values of are: \(X=0,~1,\text{ or }2\).
The probability function is \(\mathcal{P}(X=x)=\dfrac{\dbinom{2}{j}\cdot\dbinom{10}{3-x}}{\dbinom{12}{3}}\). SEE HERE

 

[Steven G]

How on earth are the three you pick going to have 0 spoiled pears AND 1 spoiled pear. It is not possible that they can happen at the same time. So what is p( A and B) equal to if A and B can NOT happen at the same time?

To do probability problems you have to know what AND means!!

 

[Loganblahtimes2]

I'm sorry for not understanding, I have the feeling I'm missing something crucial and simple but I'm not grasping it. P=(0 OR 1). I'm also not trying to find the probability of picking 0 or 1 spoiled pears, I'm trying to find how many ways I can pick 3 pears if 0 or 1 of the pears picked is spoiled. Just restating that. If A and B can not happen at the same time, P=(0 or 1).

 

[pka]

I'm sorry for not understanding, I have the feeling I'm missing something crucial and simple but I'm not grasping it. P=(0 OR 1). I'm also not trying to find the probability of picking 0 or 1 spoiled pears, I'm trying to find how many ways I can pick 3 pears if 0 or 1 of the pears picked is spoiled. Just restating that. If A and B can not happen at the same time, P=(0 or 1).

Have you studied reply #22?

 

[Steven G]

You are trying to find p(if you pick 3 pears that you get 0 or 1 spoiled pear) = p( you pick three pears and exactly 0 are spoiled OR you pick three pears and exactly 1 is spoiled)

Let A = you pick three pears and exactly 0 are spoiled and B = you pick three pears and exactly 1 is spoiled

Now using the letters A and B you are correct that you want to find p( A or B) BUT p( A or B) = p (A) + p(B) - p(A AND B). So in order to find out what you want you need to find 3 different probabilities, namely p(A), p(B) and p (A and B).
So please compute:
p(A)
p(B)
p(A and B)

 

[Steven G]

Have you studied reply #22?

It is hard for the OP to read the combinations

 

[pka]

It is hard for the OP to read the combinations

Then why ask a question that require their use?

 

[Loganblahtimes2]

Have you studied reply #22?

Yes. I changed my answer. I'm alright and thanks for helping.
Alright. If I'm reading this correctly, A=picking 3 and 0 are spoiled. B=picking 3 and 1 is spoiled. I thought that A=0, B=1 and just that. I didn't know choosing was a part of it.
If I do (3 nCr 0) it equals 1. If I do (3 nCr 1) it equals 3. Would p(A and B) then be p(1 3)?

 

[Loganblahtimes2]

Yes. I changed my answer. I'm alright and thanks for helping.
Alright. If I'm reading this correctly, A=picking 3 and 0 are spoiled. B=picking 3 and 1 is spoiled. I thought that A=0, B=1 and just that. I didn't know choosing was a part of it.
If I do (3 nCr 0) it equals 1. If I do (3 nCr 1) it equals 3. Would p(A and B) then be p(1 3)?

Sorry, I meant I changed my reply for #21. I don't understand what J would equal in #22

 

[pka]

Sorry, I meant I changed my reply for #21. I don't understand what J would equal in #22

SORRY. It should read.
Now, the possible values of are: \(X=0,~1,\text{ or }2\).
The probability function is \(\mathcal{P}(X=x)=\dfrac{\dbinom{2}{\bf x}\cdot\dbinom{10}{3-x}}{\dbinom{12}{3}}\). where \(\bf x=0,1,2\) SEE HERE
If you click on the link and scroll down to the result section you will see the table giving the probabilities of no, one, or two spoiled pears of three.

 

[Steven G]

Let's look at your first answer of 1320. I do not agree. Let's label the pears a, b, c, d, e, f, g, h, i, j, k, m, and n.

If the order in which you draw your sample matters, then you are correct that the number of possible samples is

[MATH]\dfrac{12!}{(12 - 3)!}[/MATH].

But in fact you do not care what order you select a, b, and c. That represents

[MATH]3![/MATH] of each ordered sample.

So your first answer should be [MATH]\dfrac{12!}{(12 - 3)! * 3!} = \dfrac{1320}{6} = 220.[/MATH].

Ah Jeff, you need to relearn your alphabet as n is the 14th letter so if you leave out L you have 13 letters, not 12. I only know this because when I got my first teaching job I lived in an apartment in building N which was building #14

 

[Steven G]

Yes. I changed my answer. I'm alright and thanks for helping.
Alright. If I'm reading this correctly, A=picking 3 and 0 are spoiled. B=picking 3 and 1 is spoiled. I thought that A=0, B=1 and just that. I didn't know choosing was a part of it.
If I do (3 nCr 0) it equals 1. If I do (3 nCr 1) it equals 3. Would p(A and B) then be p(1 3)?

Picking three pears and getting EXACTLY 0 SPOILED PEARS AND EXACTLY 1 SPOILED PEAR = { } (the empty set). After all if your e pears had 0 spoiled pears how can it possibly have 1 spoiled pear. Now the p( { } ) = probability of the empty set = 0. Always!

So in our case p( A or B) = p(A) + p(B) + p( A and B) = p(A) + p(B).

So all you have to do is compute p(A) and p(B) and add them together. Give it a try!

 

[Loganblahtimes2]

Picking three pears and getting EXACTLY 0 SPOILED PEARS AND EXACTLY 1 SPOILED PEAR = { } (the empty set). After all if your e pears had 0 spoiled pears how can it possibly have 1 spoiled pear. Now the p( { } ) = probability of the empty set = 0. Always!

So in our case p( A or B) = p(A) + p(B) + p( A and B) = p(A) + p(B).

So all you have to do is compute p(A) and p(B) and add them together. Give it a try!

Before I did (3 nCr 0) = 1 and (3 nCr 1) = 3. Adding those together would be 4.

 

[JeffM]

Ah Jeff, you need to relearn your alphabet as n is the 14th letter so if you leave out L you have 13 letters, not 12. I only know this because when I got my first teaching job I lived in an apartment in building N which was building #14

Off to the korner

 

[Steven G]

Before I did (3 nCr 0) = 1 and (3 nCr 1) = 3. Adding those together would be 4.

Yes, (3 nCr 0) = 1 and (3 nCr 1) = 3. So are you saying that the answer is 4?