A set S is composed of 20 multi-colored objects

[Abaddon_ire]

There are three types of people: those you can count, and those who can't.

Probabilities are all about counting.

---

I have a probability question that I need to solve:

A set S is composed of 20 multi-colored objects. 4 objects were chosen randomly from S.
What is the probability that 3 of the chosen objects be blue; knowing that there are 8 blue objects in S ?


So, I don't need hell loads of math theory. I only want steps for a solution and I'll be thankful.
Please help me.

OK, steps for a solution:

1. Determine how many ways there are for picking out three blue objects.
2. Determine how many different ways there are to pick out four objects.
3. Divide the result from 1 by the result from 2.

I am wondering what exactly is being tested here. Basic probability or combinations and permutations at a level up from that? OP did not specify.

I framed my response on pragmatism. For such a small number of elements, it quicker to do some rough and ready quick calculations. For larger sets that would be utterly impractical.

If the OP was seeking to go down the nCk/nPk route, I might have framed it differently.

Still wouldn't have given a complete answer, though. It ain't our job to do his homework.

[Mr Greene]

He also failed to say whether he wanted a ratio or percentage, and that's even assuming he got the number of blues/picks correct. (given the difference I'm not convinced)

[Abaddon_ire]

He also failed to say whether he wanted a ratio or percentage, and that's even assuming he got the number of blues/picks correct. (given the difference I'm not convinced)

I suppose we can only take the problem as posited, unless the OP clarifies. There are plenty of solutions from simple enumeration on up. And plenty of ambiguities, such as, is a chosen object replaced or removed entirely from the set of choices?

[Mr Greene]

In which case I would assume that the 4th selection would be non-blue as if the colour doesn't matter the answer is 1.

[The Valkyrie]

It's like a foreign language...

[Abaddon_ire]

In which case I would assume that the 4th selection would be non-blue as if the colour doesn't matter the answer is 1.

Within the parameters given, the answer cannot be 1. The colour matters by the definition of the question. 

If one were tasked with picking 4 objects from a set of 20 objects, then 1 would be a reasonable answer for the probability. Anyone could pick 4 from an arbitrary group of twenty objects given no further criteria. That is, however, not the task at hand. 

It is true that if one has already picked three blues, then the choice of the fourth is irrelevant at that point since three blues are already chosen. However, that is not the only possible scenario.

One cannot simply select a preferred scenario arbitrarily out of many scenarios and declare that to be the only scenario.

As posed, the colour DOES matter, since it matters whether one reaches into the imaginary bag/box/container/whatever and retrieves either a blue object or a not blue object.

If one selects the scenario that three blue objects have already been chosen, then the fourth object is irrelevant. Three blue objects have been chosen already. The criteria are satisfied. It matters not what colour the fourth one is.

But what if the first one is yellow, or red, or green, or any other colour?

Surely you can see that there are more possibilities, no?

[Mr Greene]

In which case I would assume that the 4th selection would be non-blue as if the colour doesn't matter the answer is 1.

Within the parameters given, the answer cannot be 1. The colour matters by the definition of the question. 

If one were tasked with picking 4 objects from a set of 20 objects, then 1 would be a reasonable answer for the probability. Anyone could pick 4 from an arbitrary group of twenty objects given no further criteria. That is, however, not the task at hand. 

It is true that if one has already picked three blues, then the choice of the fourth is irrelevant at that point since three blues are already chosen. However, that is not the only possible scenario.

One cannot simply select a preferred scenario arbitrarily out of many scenarios and declare that to be the only scenario.

As posed, the colour DOES matter, since it matters whether one reaches into the imaginary bag/box/container/whatever and retrieves either a blue object or a not blue object.

If one selects the scenario that three blue objects have already been chosen, then the fourth object is irrelevant. Three blue objects have been chosen already. The criteria are satisfied. It matters not what colour the fourth one is.

But what if the first one is yellow, or red, or green, or any other colour?

Surely you can see that there are more possibilities, no?

The scenario specifies 3 blue objects are picked. If the fourth can be any colour including blue then no matter what position the 'any-colour' object is picked the chances are 20/20, 19/19, 18,18 or 17/17. in any case the the answer is 1 and multiplying that by the chances for the blue selections would be a pointless exercise.
The extra selection must therefore be 'not-blue' if the scenario has any purpose.
This would insert a (12/X) selection into the matrix with four permutations.
It seems similarly pointless if the object is returned to the pool before next draw.

[Abaddon_ire]

<deleted.> I comprehensively borked the quotes by means of letting the tags escape. Let me try again.

[Abaddon_ire]

The scenario specifies 3 blue objects are picked.

Nope. The scenario specifies four are picked. but only the most suitable are selected.

If the fourth can be any colour including blue

As specified, it matters not what the fourth colour is.

then no matter what position the 'any-colour' object is picked the chances are 20/20, 19/19, 18,18 or 17/17. in any case the the answer is 1 and multiplying that by the chances for the blue selections would be a pointless exercise.

False. Actually obviously false. Only 12 out of twenty can be "not-blue". To claim 50/50 and then claim claim that all objects are blue or red is nonsense, which immediately torpedoes you argument. You are flying at that point, in the face of one's own argument.

The extra selection must therefore be 'not-blue' if the scenario has any purpose.

Nope. They decision of blue or not blue is arbitrary. Arbitrary decisions have no purpose by definition.

This would insert a (12/X) selection into the matrix with four permutations.

12x? I would dearly like to consider that a typo. I hope so, but I fear not.

It seems similarly pointless if the object is returned to the pool before next draw.

Rather meaningless. Nobody gets "returned" to anything. There is no evidence of anyone being returned to anything. Why would any rational person accept anyone accept anything about a person being returned from anything to anywhere?

And bear in mind that I am an OOBE witness, an NDE witness, through personal experience, so please do not wheel out the usual baloney.

[Paleophyte]

I have a probability question that I need to solve:

A set S is composed of 20 multi-colored objects. 4 objects were chosen randomly from S.
What is the probability that 3 of the chosen objects be blue; knowing that there are 8 blue objects in S ?

Zero. The question clearly states that "A set S is composed of 20 multi-colored objects." None of them are blue, so it doesn't matter how many of them you pick.

Better question: If you try and get the internet to do your homework for you what are the odds that one of the replies is actually your mom's phone number?