Elementary event more likely to happen-lost

hello

1.
Sample sample has 20 possible outcomes. Out of those 20 four outcomes are elements of event A. So P(A) = 4/20


But we could have similar situation, with one difference: there are 16 outcomes in sample space

P(1) = P(2) = ... = P(15), but P(16) = 4*P(1).

So P(16) is four times more likely to occur than any other elementary event. My interpretation of this particular sample space is that there are actually 20 elementary events, of which the last four make an event A. And so again, P(A) = 4/20.

-But I fear that my interpretation of an outcome 4 times more likely to happen than other outcomes, is very inflexible and won't help me much when examples get more and more complicated.

-Example of this would be a crooked die, where chances of up face being having value 3 are greater than chances of up face having any other value. In this case my interpretation doesn't seem very practical?

-So can you tell me how you interpret in your mind outcomes that are more likely to happen than others?

-Or is my interpretation of probability formula

P(A) = number of ways event A can occur/ The Total Number Of Possible Outcomes

totally correct and should always interpret it this way?



2.
P(B) = 3 / 20. Is correct way to think of this probability that out of 20 experiments of throwing a die 3 outcomes will be event B?

Or is it that chances of event B happening at each experiment are 1/20?

The latter one doesn't seem very logical, since if one has already thrown a die 15 times and and event B happened 3 times, than chances of the next toss of a die being an event B are less than 3/20?


3.
Set has 2^n subsets. For this reason we can make 2^n different sums of events ( one of those events is also impossible event N ) from sample space containing n elementary events. What kind of subset is impossible event? One with no elements in it? As I see it that's just empty set, not impossible event.

thank you for your time

 

On Mon, 10 Apr 2006 18:19:48 EDT, something_about_math

You need to put occasional line feeds in your posts.

That is correct, assuming all outcomes are equiprobable. So each point
in the sample space has probablility of 1/20.

You probably mean:
P(1) = P(2) = .. P(15) = 1/20 and P(16) = 5/20 since the probabilities
must add to 1.

No, you shouldn't interpret this as you describe. You have a sample
space with 16 outcomes. They just aren't equiprobable. For example,
they could represent the outcomes of a spinner where the 16 area is
larger than the others.

You will learn that many useful sample spaces are exactly of the type
where the outcomes are not equally probable.

Like you observed, a biased die is a good example.

Say you have a sample space with n outcomes e_1, e_2, ...e_n, where
the probability of the i'th outcome is p_i. All you need is for the
p_i >= 0 and sum p_i = 1. They don't have to be equal.

Then if you have an event E = {e_2, e_10, e_12}, to compute the
probability of E you just add the probabilities of the outcomes in E.
In this example, P(E) = p_2 + p_10 + p_12.

It is *only* correct for uniform sample spaces where all outcomes are
equally probable.

No.

It means that on any trial, the probability of B is 3/20.
Statistically, if you repeated the experiment many times the
proportion of times event B occurs will be approximately 3/20, but not
necessarily exactly. The more times the experiment is repeated, the
closer the ratio will tend.

You're right, it isn't logical.

Not "sums of events". Just 2^n events. They *are* subsets.

The empty set is { }. Its probability is zero.

--Lynn

 

by my count 15+4*1 is 19

but you said four times more likely, which means 5 times as likely.

so 15 + 5 is which is 20 as you say.

generally in statistics probabilities are given as fractions that add up to 1
when all possible outcomes are considered.

this is only good when each outcome is equally likel, like when using fair
dice...

you could consider it like a fair 20 faced die with three faces marked with B.

but really all it is is a fraction 3/20 or 0.15 , b will occur 0.15 of the
time (or 15% of the time)

that seems confused.

if the impossible event N is in the starting set,
the impossible event N will be in half of the subsets,

any subset that includes the impossible event will also be impossible.

an impossible event is just an event with a probability of 0

consider the set { R = it rains today , M = I get mail, N = pigs fly }

the wether man tells me there's 30% chance of rain and I get mail 1 day in 5
but pigs never fly, probability 0.

so the subsets are

{}
(N}
{M}
{M,N}
{R}
{R,N}
{R,M}
{R,M,N}

all the ones containing N have probability 0, the empty set however
(meaning no rain, no mail, and no flying pigs) is quite possible.

Bye.
Jasen

 

-Example of this would be a crooked die, where

I understand why it must add up to 1 when all
outcomes are equiprobable, but not why up to
1 when one outcome is 4 times more likely to happen
than others!

Out of all tosses?

Please elaborate.

Why half? :0

Why will it be impossible? If subset has
E:{pigs fly, it rains today}, then E happens if
either it rains or pigs fly. Since pigs can't fly E
happens when it rains:

pigs_fly U it_rains_today = it_rains_today

 

2.

This may seem utterly stupid question, but when you
say on any trial ( which in this case means one throw
of a die ) the probability is 3/20, what does that mean? It really bugs me.

It isn't logical and yet it is correct?

If my interpretation isn't correct then I must admit
that I have no clue why probabilities must add to 1,
since P(16) is considered on "entity" out of 16 total!

 

On Tue, 11 Apr 2006 21:04:54 EDT, something_about_math

Then in fact you don't understand why they must add up to 1
when the outcomes are equiprobable either, because the
reason is the same in all cases. Basically it's a matter of
definition. Probability is defined so that (a) if events
E_1, E_2, ..., E_n are pairwise disjoint (i.e., no two of
them can occur simultaneously), then

P(E_1 or E_2 or ... or E_n) =
P(E_1) + P(E_2) + ... + P(E_n),

and (b) if U is the event consisting of all possible
outcomes, then P(U) = 1. (Loosely speaking, if an event is
certain to occur, its probability is 1.) From these it
follows immediately that the probabilities of the individual
outcomes must sum to 1: simply consider each outcome to be
an event, and note that the outcomes are pairwise disjoint.

Consider a die with one red face and five blue faces. The
faces are otherwise unmarked, so you can't tell one blue
face from another. When you roll the die, there are only
two possible outcomes: it lands with red up, or it lands
with blue up. Obviously it's more likely to land with blue
up than with red up; in fact, blue is five times as likely
as red: P(red) = 1/6, and P(blue) = 5/6.

If you toss it many times, B will occur about 15% of the
time; the more times you toss it, the closer to 15% the
proportion of B's is likely to be.

It's really no different from tossing a fair coin. The
probability that it comes up heads is 1/2, but you surely
must realize that if you toss the coin 10 times, you may
very well *not* get exactly 5 heads. Indeed, if you toss
the coin 1,000,000 times, you are very, very unlikely to get
exactly 500,000 heads. But your *percentage* of heads is
very unlikely to be less than 49.985% or more than 50.015%.

The tosses are completely independent. What happens in the
first 15 tosses has absolutely no effect on the probability
of getting B on the 16th toss.

[...]

Brian

 

something like that, it depnds what you mean by all tosses...

I can't follow your loggic.

it's in half the unique subsets, like any other element...
it's the reciprocal of the 2 in 2^.

I guess it depends on what you take set, its elements, and
the subsets to mean, and I appear to have misinterpreted it.

Bye.
Jasen

 

2.

So how is that different from me interpreting that
3/20 means that if we toss a die with with number 2
three times more likely to happen, then the desired
face up will happen around 3 times if we toss it
20 times?

So chances of of tossing 10 heads and prob of
tossing 10 heads after you have already tossed
9 heads are equal?

 

So set has an impossible event only if we include
it in a set.

If an experiment is to roll a die, and if outcomes
are

{1,2,3,4,5,6}

then this set doesn't have an imposible event.

But if outcomes are

{1,2,3,4,5,6,7}

then set has impossible event - > 7 ?

 

On Wed, 12 Apr 2006 21:04:21 EDT, something_about_math

If you toss the thing 20 times, it would not be terribly
surprising to get the B face only once, or four or five
times. If *many* of us toss it 20 times each, however, and
we average the numbers of times we get the B face, that
average is very likely to be very close to 3.

Yes. You're starting from scratch with each toss.

Brian

 

On Wed, 12 Apr 2006 21:05:18 EDT, something_about_math

Yes, if it's an ordinary cubical die. And the seven
outcomes are of course not equally likely; their
probabilities are 1/6, 1/6, 1/6, 1/6, 1/6, 1/6, and 0, in
order.

Brian

 

You are not thinking about this in the right way - probably you are muddling
"events" and "individual outcomes" for your trials...

You have said the possible outcomes of an experiment (trial)
are 1,2,3,4,5 and 6, i.e. the elements of the set A = {1,2,3,4,5,6}. Then
"events" that you can look at will be all the different *subsets* of A.

E.g. there is an event "outcome is 1", which is the event {1},
or "outcome is even" which is {2,4,6}
or "outcome is odd" which is {1,3,5}
or "outcome is even or odd" which is {1,2,3,4,5,6} = A
or "outcome is more than 4" which is {5,6}
or "outcome is even and odd" which is {} = empty set
etc. etc.

Every event is a subset of A, and every subset of A is an event. How many
different events can we concoct out of set A above? Since each number 1-6
can be either included or excluded, there are 2*2*2*2*2*2 = 64 events.

Remember that all sets contain themselves as a subset, and contain the empty
set as a subset. In terms of events, this means there will always be the
following events:

{} (empty set) = "impossible event", which has probability 0
A (full set) = "certain event", which has probability 1.

The empty set has probability zero, because if you perform a trial you are
certain to get an element of A (by definition), and this element is not in
the empty set - so the "impossible event" has not occured.

The set A has probability one, because whichever outcome occurs, it is (by
definition of A) in set A, i.e. the event A has occured.

Hope this helps,
Mike.