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Is anyone going to help?

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The Cubs are playing the Red Sox in the World Series. To win the world series, a team must win 4 games before the other team does. If the Cubs win each game with probability 3/5 and there are no ties, what is the probability that the Cubs will win the World Series? Express your answer as a percent rounded to the nearest whole percent.

Apr 13, 2018

#1
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how many matches does each team play in total?   Apr 13, 2018
edited by lynx7  Apr 13, 2018
#2
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it  only says that they need to win 4 games before the other team

MIRB16  Apr 13, 2018
#3
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3/5 + 3/5 + 3/5 + 3/5 =12/5 (probability that Cubs will win all four matches)

12/5 * 100 = 240%

i am not sure if i am right.   Apr 13, 2018
#4
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If the probability of winning any one game is 3/5, then the probability of winning the World Series should be: (3/5)^4 =0.1296 =~13%.

Apr 13, 2018
#5
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I think the probability should be:

7C4 x (3/5)^4 x (2/5)^3 =~29%

Apr 13, 2018
#6
+2

Let everybody chime in!. I think you have to sum up the probabilities as follows:

[7C4 x 0.6^4 x 0.4^3] + [7C3 x 0.6^5 x 0.4^2] + [7C2 x 0.6^6 x 0.4^1] + [7C1 x 0.6^7 x 0.4^0] =

0.290 304               +        0.261 273 6         +        0.130 636 8        +        0.027 993 6        =0.710208 =~71%.

Apr 14, 2018
#11
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Guest #6 is Mr. BB.  He’s a semi-skilled plagiarist, and he tends to copy and paste a posted answer/solution that’s defective.   Of course, there is no explanation for its derivative, and in this case, the binomials are wrong. The counts should be from 4 to 7, not 4 to 1.  When Mr. BB serves porkbutt roasts the anuses are still attached. Nothing too unusual, here. GA

GingerAle  Apr 14, 2018
#7
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I believe this is the correct answer:

$\sum_{k=4}^{7}\binom{7}{k}0.6^k\cdot{0.4}^{7-k}$

The gain is that it becomes a routine binomial distribution with fixed $n$.

I hope this helped!

Apr 14, 2018
#8
+1

That gives exact same result as Guest #6's answer!.

Guest Apr 14, 2018
#10
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Yes, that is the correct answer.

sum(nCr(7, k)*0.6^k*0.4^(7-k),k,4,7)

Enter this into the forum’s calculator for a confirmation.

If you are going to paste LaTex code on here, you should also open a LaTex box and paste it in there so that it renders. You should also give a source for your code!

You may hope it helps, but it does not help very much.

While this is the correct result, without an explanation of its origination, it’s of little help to a student trying to learn how to solve these types of probability questions.

GA

GingerAle  Apr 14, 2018
#9
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Solution:

$$\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} (3/5)^4 = (81/625) = 12.96\%\\ \text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \binom{4}{3}(3/5)^3*(2/5) *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \binom{5}{3}(3/5)^3*(2/5)^2 *(3/5) = (648/3125) \approx 20.74\% \\ \text{ }\\ \text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games in six trials and then win the seventh game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 6})* \rho \text{(Cbs win 7th game}) =\\ \text{ } \hspace{34 mm} \binom{6}{3}(3/5)^3*(2/5)^3 *(3/5) \approx 16.59\% \\ \text{ }\\ \text{ The sum of the individual probabilities gives }\\ \text{ the overall probability of the Cubs winning the series.}\\ \text {Sum of individual probabilities: } \\ (20.74+20.74+16.59+12.96) \bf \approx 71.0\%$$

GA

Apr 14, 2018
#12
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Thank you All!

Apr 14, 2018