Solution to Problem 2

A Math Forum Project

Table of Contents:

Famous Problems Home

The Bridges of Konigsberg
· Euler's Solution
· Solution, problem 1
· Solution, problem 2
· Solution, problem 4
· Solution, problem 5

The Value of Pi
· A Chronological Table of Values
· Squaring the Circle

Prime Numbers
· Finding Prime Numbers

Famous Paradoxes
· Zeno's Paradox
· Cantor's Infinities
· Cantor's Infinities, Page 2

The Problem of Points
· Pascal's Generalization
· Summary and Problems
· Solution, Problem 1
· Solution, Problem 2

Proof of the Pythagorean Theorem

Proof that e is Irrational

Book Reviews

References

Links

Finding the probability that Pascal will be ahead 9 to 6 amounts to finding the probability that, in 5 flips of the coin, Pascal will win exactly 4 times. This is given by

5C4
-----------------------------------------------
total possible outcomes of 5 coin flips

Now, the number of ways you can choose r objects from five objects, for different r, is given by the 6th row of Pascal's triangle, which is

1  5  10  10  5  1

Thus the total number of outcomes is 1+5+10+10+5+1 = 32. And, 5C4 = 5. So the probability that Pascal will be ahead 9 to 6 is 5/32 = 0.15625. That is, we expect that this will happen 15.625 percent of the time.

 

 

to Probability: Summary and Problems
to a Proof of the Pythagorean Theorem

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August, 1998