Probability of Uncorrelated Events

Dear Experts:

I have what I think are a couple of basic probability questions that I can't get my head around.  I don't know anything about statistics, so bear with me.  

Here's my question:  assume that are four independent, uncorrelated events, each of which has a 25% chance of occurring.  Three questions:

1)  What is the chance that any one of the four events will occur?  In other words, there are four 25% chances--is there thus a 100% probability of one of the events occuring?

2)  What is the probability that any two of the four events will occur?

3)  Is the probability that all four events will occur equal to (.25 x .25 x .25 x .25)?

Please excuse if I am using improper or inexact terminology.  Please include the basic formula with any answer...

Regards,

TMR
tmreiterAsked:
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JR2003Commented:
four independent, uncorrelated events A, B, C, D

1)  What is the chance that any one of the four events will occur?  In other words, there are four 25% chances--is there thus a 100% probability of one of the events occuring?
This is the chance of
A happening and not B, C or D
+
B happening and not A, C or D
+
C happening and not A, B or D
+
D happening and not A, B or C

This is
(0.25*0.75*0.75*0.75)+(0.25*0.75*0.75*0.75)+(0.25*0.75*0.75*0.75)+(0.25*0.75*0.75*0.75)
=0.421875



2)  What is the probability that any two of the four events will occur?
There are 6 way that exactly 2 events will occur:
ABCD
------
0011
0101
0110
1001
1010
1100

=(0.75*0.75*0.25*0.25)*6
=0.21093756

3)  Is the probability that all four events will occur equal to (.25 x .25 x .25 x .25)?
Yes you care correct
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aburrCommented:

Question one. There are four possible events each with a 25% possibility. An event happens. Of course there is 100% chance of one of the four events happening. There is no other possibility.
You have fourr cards (the four aces.) Each ahs a 25 % chance of being picked. You pick a card. Of course one of the aces has been picked.

Question two. You pick two cards What is the probablity of two cards being pick 100% of course. I think that you really wanted to ask a different question.

Question three. You pick four cards. What is the probablity that all the cards will be picked. Again 100% You had only foru cards so of course all are picked. Again I think you wanted to ask a different question but I am not sure what that question is.

You may want to replace the card after it is picked? You may want to ask something else.
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tmreiterAuthor Commented:
Dear aburr:

Yes, we seem to be addressing different questions.  Let's stick with the four cards example.  Let's say that in Question 1, there are four cards on the table.  Each card could be a heart, diamond, club, or spade (and if card #1 is a particular suit, it does not affect the probability of any of the subsequent cards being one suit or another).   So now the question is what is the probability that at least one of the four cards will be a spade.  Same scenario for Question 2, but what is the probability of two of the cards being spades?  Hope this is more clear!

TMR
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ozoCommented:
For uncorrelated events, you multiply the probabilities.
The probability of an event not happening is .75
the probability of 4 events not happening is (.75 x .75 x .75 x .75)
the probability of at least one of the events happening is 1-(.75 x .75 x .75 x .75)
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aburrCommented:
I now understand your question better.
The difference between the answer (to question1) between ozo and JR2003 is that JR2003 gave the probabliity of exactly one event happening while ozo gave the probablity of at least one happening. His probability included two, three, and four happenings.
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jitendra_wadhwaniCommented:
>>1)  What is the chance that any one of the four events will occur?  In other words, there are four 25% chances--is there thus a 100% probability of one of the events occuring?

Means only one should occur and remaining should not occur

Only A  =  (1/4)*(3/4)*(3/4)*(3/4)
Only B  =  (1/4)*(3/4)*(3/4)*(3/4)
Only C  =  (1/4)*(3/4)*(3/4)*(3/4)
Only D  =  (1/4)*(3/4)*(3/4)*(3/4)

Any of them Add All = (3/4)*(3/4)*(3/4)



Ohh.. JR2003 has already given all answers

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tmreiterAuthor Commented:
Guys, thanks for the responses--sorry for the delay in getting back to you, but I've been out of town.  I think JR2003 answered the questions best, so I'll give him the points.

TMR
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