Question 187358This question is from textbook Survey of Mathematics w/ Applications expanded edition
: I'm totally lost with the probability and statistics, I have 2 discussions questions and don't know where to start.
1.) Explain using the definition of probability why: A) the probability of an event that cannot occur is 0. B) The probability of an event that must occur is 1.
2.) How do I explain permutations and combinations and the difference between the two, and what are some examples to illustrate the 2.
This question is from textbook Survey of Mathematics w/ Applications expanded edition
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1.) Explain using the definition of probability why:
A) the probability of an event that cannot occur is 0.
P(success) = (# of ways success can happen)/(# of ways to succeed + # of ways to fail)
If the event cannot occur the # of ways to succeed is 0. Then the numerator
of the probability fraction is zero. Then the probability is zero.
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B) The probability of an event that must occur is 1.
If it must occur the # of failures is zero.
So P(success) = (# successes)/(# successes + 0) = (# successes)/(# successes)=1
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2.) How do I explain permutations and combinations and the difference between the two,
# of permutations of n objects is the # of ways to arrange n objects:
n ways to choose position #1
n-1 ways to choose position #2
n-2 wya to choose position #3
etc.
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# of combinations of n object is the number of distinct groups of size r
that can be formed using the n objects.
The number of ARRANGEMENTS of r of the n objects is n*n-1*n-2*...*(n-r+1)
But each set of r of these objects appears r*r-1*r-2*...*1 times is that
large set of arrangements.
So the # of unique sets of r of the n objects is (# of arrangements of the
object) divided by (# of arrangements of r of the objects)
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and what are some examples to illustrate the 2.
Say you have three objects: a,b,c
# of arrangements of all three is 3*2*1 = 6
# of combinations of all three is 6/6 = 1
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# of arrangements of 2 of the objects is 3*2 = 6
# of combinations of 2 of the 3 objects is 6/(2*1) = 3
{a,b},{a,c} and {b,c} are the three combinations.
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Cheers,
Stan H.
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