MHB Kate's Chances of Winning a 5 Match Tennis Tournament

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Kate has an 80% probability of winning each tennis match. In a 5-match knockout tournament, her chance of winning the tournament is calculated as 0.8^5, resulting in approximately 33%. If she wins her first three matches, the probability of winning the tournament remains the same, as winning all five matches includes winning the first three. The discussion clarifies that the intersection of winning the first three and winning all five is simply winning all five matches. Therefore, the probability calculations reinforce that winning all five matches is the only way to satisfy both conditions.
Bushy
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The Probability of Kate winning a tennis match is 80%. If she enters a 5 match knockout tournament, find the chance of her:

a) winning the tournament:


Is simply 0.8^5 = 33%

b) winning the tournament given she wins her first three games:

Is the intersection of the two over the probability she wins the first three = 0.8^5 * 0.8^3 / 0.8^3 = the same as part a?
 
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Bushy said:
The Probability of Kate winning a tennis match is 80%. If she enters a 5 match knockout tournament, find the chance of her:

a) winning the tournament:


Is simply 0.8^5 = 33%

Keep your answer exact please. $\displaystyle \begin{align*} \left( \frac{4}{5} \right) ^5 = \frac{1024}{3125} \end{align*}$.

b) winning the tournament given she wins her first three games:

Is the intersection of the two over the probability she wins the first three = 0.8^5 * 0.8^3 / 0.8^3 = the same as part a?

No, winning all five and winning the first three is equivalent to simply winning all five. So the intersection is simply what you found in part (a).
 
Prove It said:
No, winning all five and winning the first three is equivalent to simply winning all five. So the intersection is simply what you found in part (a).

Therefore 0.8^3 / 0.8^3 = 1 ?
 
Bushy said:
Therefore 0.8^3 / 0.8^3 = 1 ?

NO! The top is the probability of winning all FIVE!
 
Well that would give a neat answer, but aren't we saying the intersection of winning the first three and winning all five should be winning the 1st three?
 
Bushy said:
Well that would give a neat answer, but aren't we saying the intersection of winning the first three and winning all five should be winning the 1st three?

No, the only way it is possible to do both "winning all five games" and "winning the first three games" is to win all five games!
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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