Hey
If the odds in favor of Boris beating Elena in a chess game are 5 to 4, what is the probability that Elena will win an upset victory in a best-of-five chess tournament?
Can someone give me an answer with explanation Plz.
Thanx
Hey
If the odds in favor of Boris beating Elena in a chess game are 5 to 4, what is the probability that Elena will win an upset victory in a best-of-five chess tournament?
Can someone give me an answer with explanation Plz.
Thanx
Hello, ballin_sensation!
The probability of Elena winning a game is $\displaystyle \frac{4}{9}$ . . . of losing, $\displaystyle \frac{5}{9}$If the odds in favor of Boris beating Elena in a chess game are 5 to 4, what is the
probability that Elena will win an upset victory in a best-of-five chess tournament?
There are three ways in which she can win three games.
(1) She wins the first three games:
. . . $\displaystyle \left(\frac{4}{9}\right)^3 \:=\:\frac{64}{9^3}$
(2) She wins two of the first three games, then wins the fourth game.
. . . $\displaystyle {3\choose2}\left(\frac{4}{9}\right)^2\left(\frac{5 }{9}\right)\cdot\frac{4}{9} \;=\;\frac{960}{9^4}$
(3) She wins two of the first four games, then wins the fifth game.
. . . $\displaystyle {4\choose2}\left(\frac{4}{9}\right)^2\left(\frac{5 }{9}\right)^2\cdot\frac{4}{9} \;=\;\frac{9600}{9^5}$
Therefore: .$\displaystyle P(\text{Elena wins}) \;=\;\frac{64}{9^3} + \frac{960}{9^4} + \frac{9600}{9^5} \:=\:\frac{23,424}{59,049} \;=\;0.396687497 \;\approx\;40\%$
This seems like a binomial probabilility so
the prob of x successes in n trials is
$\displaystyle P(x)=\binom{n}{x}P^{x}(1-P)^{n-x}$
where P is the prob of success
since we want an upset victory she needs to win 3,4, or 5 games
so we can add P(3) +P(4)+P(5) or take 1 -P(0)-P(1)-P(2)
Now lets set up the formula:
The odds are 5 to 4 against her so $\displaystyle P=\frac{4}{9}$
and they are playing the best of 5 so
$\displaystyle P(x)=\binom{5}{x}\left( \frac{4}{9}\right)^{x} \left(\frac{5}{9} \right)^{5-x}$
$\displaystyle P(3)=\binom{5}{3}\left( \frac{4}{9}\right)^{3} \left(\frac{5}{9} \right)^{5-3}=\frac{5!}{3!2!} \cdot\left( \frac{4}{9}\right)^{3} \left(\frac{5}{9} \right)^{5-3}= 10 \frac{(4^3)(5^2)}{9^5} \approx .2710$
$\displaystyle P(4)=\binom{5}{4}\left( \frac{4}{9}\right)^{4} \left(\frac{5}{9} \right)^{5-4}=\frac{5!}{4!1!} \cdot\left( \frac{4}{9}\right)^{4} \left(\frac{5}{9} \right)^{5-4}= 5 \frac{(4^4)(5^1)}{9^5} \approx .1083$
$\displaystyle P(5)=\binom{5}{3}\left( \frac{4}{9}\right)^{5} \left(\frac{5}{9} \right)^{5-5}=\frac{5!}{5!0!} \cdot\left( \frac{4}{9}\right)^{5} \left(\frac{5}{9} \right)^{5-5}= 1 \frac{(4^5)(5^0)}{9^5} \approx .0173$
so P(3)+P(4)+P(5) $\displaystyle \approx .3966$