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# The Cubs are playing the Red Sox in the World Series.

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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 3/4 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.

Nov 16, 2020

#3
+2140
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Solution:

$$\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{ }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} \left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \approx 31.64%\\$$

$$\text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \dfrac{81}{256} \approx 31.64 \% \\ \text{ }\\$$

$$\text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \dfrac{405}{2048} \approx 19.78\% \\ \text{ }\\$$

$$\text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games in six trials and then win the seventh game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 6})* \rho \text{(Cbs win 7th game}) =\\ \text{ } \hspace{34 mm} \dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\dfrac{405}{4096} \approx 9.88\% \\ \text{ }\\$$

$$\text{ The sum of the individual probabilities gives }\\ \text{ the overall probability of the Cubs winning the series.}\\ \text {Sum of individual probabilities: } \\ \left(\dfrac{81}{256}\right)+ \left(\dfrac{81}{256}\right) + \left(\dfrac{405}{2048}\right)+ \left(\dfrac{405}{4096}\right) = \dfrac{3807}{4096} \bf \approx 92.94\%$$

GA

Nov 16, 2020

#1
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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 3/4 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.

Is this a trick question or just worded poorly?

Consider the third sentence:  If the Cubs win each game....

If the Cubs win each game, then the probability they win the Series is 100%.

Nov 16, 2020
#2
+2140
+1

Are you are trying to imitate Gracie Allen, or has your contagious dumbness mutated into virulent batshit stupidity?

Is this a trick question or just worded poorly?

Consider the third sentence:  If the Cubs win each game....

If the Cubs win each game, then the probability they win the Series is 100%

To answer the question –or to just understand the question, you have to read the whole sentence, Mr. BB ...you can’t just stop after the sixth word.  I know you can do this if you try....

GA

Nov 16, 2020
#4
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Oh, Ginger, such bitter hostility is all out of proportion to the transgression you imagine.  And to think, once in the past I actually defended you when you were attacked by some rude lout.  How sharper than a serpent's tooth is a thankless child.

So what if I stopped after six words.  Those words are all that were necessary.  They set the conditional clause of the sentence.  The remainer of the clause, to wit, "with probability 3/4 and there are no ties" has no practical significance.  It doesn't matter with what probability they win, because the condition that they do win is laid out in the first six words: that they win each game.  That's all we need.

I was pretty sure what the question attempted to say.  That's why I allowed for the possibility of poor composition.  At the same time I definitely knew what the question did say.  Call me a pedant if you will, but in maths one can get in trouble in a hurry if one assumes things that aren't there.

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Guest Nov 16, 2020
#6
+2140
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Ron, if you hadn’t responded to this, I’d never have known it was you. LOL

The post I trolled is definitely BB esque –it’s exactly the kind of post that at least two of the forum’s BBs would make (I can give examples).  Even though you’ve claimed ownership, the only corroborating evidence is that you included a copy of the question in your post. The BBs rarely do this.

I’ve made mistakes before, believing a guest poster to be a BB. It usually happens when the post gobbles like a turkey, walks like a turkey, and poops like a turkey; but it’s really a duck playing a turkey on the net. Before this, the only time I know for sure that the BB I trolled was an impostor is here: https://web2.0calc.com/questions/kim-has-10-identical-lamps-and-3-identical-tables#r9  EP was the imposter.  EP is no BB, but he did an excellent job of playing one on the forum.  And you did too.

While I usually enjoy busting rumps with an occasional troll post, it’s especially good when the target post is a gold nugget to a troll. This is the first time I’ve trolled you. Truly, it was a wonderful first time, but I do wish it was as good for you as it was for me.  Perhaps the next time, when I know it’s you, it will be good for both of us, and we’ll respect each other in the morning.

My favorite target is EP. Busting EP’s rump with an occasional troll post is wonderfully enjoyable. Here’s a recent one: https://web2.0calc.com/questions/helllllppppp_2#r3.  This didn’t start out as a troll post.  I just made a terse comment to Asinus.

Asinus, this is NOT the correct equation for this question.

Then EP, [Hello EP], walks half way across the Troll Bridge and jumps in the river, without any help from me. Then he swims to the river bank, dries himself off, and then walks right into my troll cave to deliver a message...  God, I laughed for ten minutes. This was two gold nuggets.

I’ve never trolled Asinus, except for a few light teases.  He’s worthy of more, but Asinus is very polite, so I’m reluctant, but still I’m tempted.

I remember when I first saw Asinus log on the forum.  When I saw his name, I laughed for an hour.  I thought, “Someone has a wonderful sense of humor or is clueless...”  Here’s why:

Sinus translates from the Latin as “half chord [of a circle]” The well known trigonometric function (that is not usually thought of in terms of half chords).  Arcsinus (also Latin), translates as “the arc that makes this half chord.” Asinus is the abbreviated form of Arcsinus. This is (obviously) Asinus’ intended meaning for his user name.

It’s relatively rare to hear or read the Latin word for “sine,” and it’s very rare to hear or read the abbreviated form “asinus,” but asinus is actually a Latin word, unrelated to the words above.  It translates to “ass” as in donkey; it also means “fool” as in jackass.  Even after six years, I still find this very funny. [LOL]

----------

Here, I speculate if you might descend into the BB realm. I didn’t think you would, but if in fact you made that post with the intent of anything other than the objective of imitating the humor of Gracie Allen, then you have.  Even so, I do not think I need to write a Requiem for Ron just yet.

Visiting the Shrine of Organized Stupidity and Perpetual Quantum Dumbness doesn’t mean you have joined The Congregation of the Brain-Dead and Damned.

Though you may have descend into the BB posting zone, I do not think that will be your new residence.  Now, the question is, “Why did you, after posting continuously for more than eighteen months, create a post that seems like you were blessed by the God of Stupid (instead of Gracie Allen)?”

I can speculate on this:

... A door to another universe is now open. Enter at your own risk... and read on to find out why and what may have compelled you to visit the BB zone.

Continued...

GA

GingerAle  Nov 21, 2020
edited by GingerAle  Nov 21, 2020
#7
+2140
0

Pending............................................

GingerAle  Nov 23, 2020
#8
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Hi Ginger, it's so good to hear from you again.  Thanks for the link to the helllllppppp_2 thread, I noted with delight your comment that "dumbness is contagious."  It reminded me of two things.  I'll start with the shorter:  Insanity is hereditary .... you get it from your children.  Parents will know what this means.

The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually.  The two hosts on the show were known for self-deprecating humor, and the question was submitted by a fan of the show.  The discussion ran for weeks, but if it was ever resolved, I missed that episode.

I almost adopted the username MisterBB for this site, in your honor, but I checked it out first on the internet and discovered to my disappointment that it had already been claimed.  Sigh... gang aft agley.

Goodnight, Gracey.

.

Guest Nov 23, 2020
#9
+2140
0

Also Pending ................

GingerAle  Dec 1, 2020
#3
+2140
+2

Solution:

$$\text {The series ends after a team has a fourth win.}\\ \text{Here are the four scenarios with derived probabilities }\\ \text{where the Cubs win the series. }\\ \text{ }\\ \text{1) The Cubs win the first four games.}\ \text{ }\\ \hspace{35 mm} \rho(\text {Cbs win on } 4^{th} \text{game )= }\ \text{ }\\ \hspace{35 mm} \left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \approx 31.64%\\$$

$$\text{ }\\ \text{2) Cubs win series in game five. }\\ \text{For the Cubs to win the series in game five,}\\ \text{ they need to win three games in four trials and then win the fifth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 4})*\rho \text{(Cbs win 5th game}) =\\ \text{ } \hspace{35 mm} \ \dbinom{4}{3}*(3/4)^3*(1/4)*(3/4)= \dfrac{81}{256} \approx 31.64 \% \\ \text{ }\\$$

$$\text{3) Cubs win series in game six. }\\ \text{For the Cubs to win the series in game six, }\\ \text{they need to win three games in five trials and then win the sixth game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 5})* \rho\text{(Cbs win 6th game}) =\\ \text{ } \hspace{34 mm} \dbinom{5}{3}*(3/4)^3*(1/4)^2 *(3/4) = \dfrac{405}{2048} \approx 19.78\% \\ \text{ }\\$$

$$\text{4) Cubs win series in game seven. }\\ \text{For the Cubs to win the series in game seven, }\\ \text{they need to win three games in six trials and then win the seventh game. }\\ \text{ } \hspace{34 mm} \rho \text {(Cbs win 3 in 6})* \rho \text{(Cbs win 7th game}) =\\ \text{ } \hspace{34 mm} \dbinom{6}{3}(3/4)^3*(1/4)^3 *(3/4) =\dfrac{405}{4096} \approx 9.88\% \\ \text{ }\\$$

$$\text{ The sum of the individual probabilities gives }\\ \text{ the overall probability of the Cubs winning the series.}\\ \text {Sum of individual probabilities: } \\ \left(\dfrac{81}{256}\right)+ \left(\dfrac{81}{256}\right) + \left(\dfrac{405}{2048}\right)+ \left(\dfrac{405}{4096}\right) = \dfrac{3807}{4096} \bf \approx 92.94\%$$

GA

GingerAle Nov 16, 2020
#10
+114094
0

.....The other thing is a concept that was discussed on the radio show Car Talk, the question whether two people together can know less than either of them individually.  .....

LOL   !!!!!

This is sometimes demonstrated on the Forum......

Dec 1, 2020