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.

MIRB16 Apr 13, 2018

#3**+1 **

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.

lynx7 Apr 13, 2018

#4**+1 **

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%.

Guest Apr 13, 2018

#6**+1 **

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%.**

Guest Apr 14, 2018

#11**0 **

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**+3 **

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!

GYanggg Apr 14, 2018

#10**+1 **

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**+3 **

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

GingerAle Apr 14, 2018