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The chart below lists the odds that a horse will place in
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17 Jun 2006, 08:23
Question Stats:
28% (02:47) correct 72% (02:57) wrong based on 403 sessions
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The chart below lists the odds that a horse will place in the top three. As part of a contest, if at least 2 of the horses that a person bets upon finish in the top three, the bettor receives a Tshirt. If Cecilia bets on The Baron, Happy Cynic, and Inamorata, what is the probability that she does not receive a Tshirt? A. 3/50 B. 21/100 C. 29/100 D. 35/100 E. 65/100
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Re: The chart below lists the odds that a horse will place in
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25 Mar 2014, 15:13
madrid2020 wrote: can someone please explain a bit more "The chart below lists the odds that a horse will place in the top three. As part of a contest, if at least 2 of the horses that a person bets upon finish in the top three, the bettor receives a Tshirt. If Cecilia bets on The Baron, Happy Cynic, and Inamorata, what is the probability that she does not receive a Tshirt?" Cecilia can win a TShirt, if at least 2 of The Baron, Happy Cynic, and Inamorata finish in top three. Now theoretically, when she will not get a TShirt ? When either only 1 or NONE of her horses can make into top three. But practically, there are total 5 horses, so anyhow 1 of her horse will make into top 3, so the NONE case is eliminated. The winning probability of the 3 horses Cecilia has bet: The Barron: 3/5 Happy Cynic: 7/10 Inamorata: 1/2 For any game with no draws: P(Win) + P(Loss) = 1 So probability that Cecilia doesn't win TShirt = Probability that only one of the above three horses win and other two loose = P(The Barron Wins)*P(Happy Cynic Looses)*P(Inamorata Looses) + P(The Barron Looses)*P(Happy Cynic Wins)*P(Inamorata Looses) +P(The Barron Looses)*P(Happy Cynic Looses)*P(Inamorata Wins) = (3/5) * (17/10)*(11/2) + (13/5) * (7/10)*(11/2) + (13/5) * (17/10)*(1/2) = 29/100
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Re: PS  Probability
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17 Jun 2006, 08:41
shobhitb wrote: The chart below lists the odds that a horse will place in the top three. As part of a contest, if at least 2 of the horses that a person bets upon finish in the top three, the bettor receives a Tshirt. If Cecilia bets on The Baron, Happy Cynic, and Inamorata, what is the probability that she does not receive a Tshirt?
1.3/50 2. 21/100 3. 29/100 4. 35/100 5. 65/100
The probability that it will not win the Tshirt=
probbility that none of the horses make it into top 3
+probability that only one of the horses that are bet make it into top 3
= 2/5*3/10*1/2+3/5*3/10*1/2+2/5*3/10*1/2+2/5*7/10*1/2
=6/100+9/100+6/100+14/100
35/100
Option 4



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probbility that none of the horses make it into top 3 should not considered hence the answer is 29/100



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gmatinjune wrote: probbility that none of the horses make it into top 3 should not considered hence the answer is 29/100
yes.you are corrrect...since there are only 5 horses one horse definitely makes it into top 3 out of 5..answer is C



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Updated on: 17 Jun 2006, 09:20
#3. 29/100 by my calculations.
Initially thought the possibility of none of the 3 horses finishing the race is a valid scenario...
Shouldn't we be factoring in the possibility of the other two horses in the top 3? In which case, the whole calculation would be multiplied with (3/4)*(1/4)... explanations?
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Originally posted by paddyboy on 17 Jun 2006, 08:51.
Last edited by paddyboy on 17 Jun 2006, 09:20, edited 1 time in total.



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Re: The chart below lists the odds that a horse will place in
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25 Nov 2013, 16:28
shobhitb wrote: Attachment: Odds.gif The chart below lists the odds that a horse will place in the top three. As part of a contest, if at least 2 of the horses that a person bets upon finish in the top three, the bettor receives a Tshirt. If Cecilia bets on The Baron, Happy Cynic, and Inamorata, what is the probability that she does not receive a Tshirt? A. 3/50 B. 21/100 C. 29/100 D. 35/100 E. 65/100 Im not sure wjere am I going wrong here. Would appreciate some guidance. We need to find the probability that either 2 or 3 finish on the top 3. So first for all of 'em to finish in the top three he have (3/5)(7/10)(1/2) Now for two of the three to finish top three we have (3C2)(3/5)(7/10)(1/2) Now by adding both we should have the total probability of winning a tshirt and By 1 P(winning) = Not winning. Is this procedure accurate? Thanks Cheers! Kudos rain J



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Re: The chart below lists the odds that a horse will place in
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26 Nov 2013, 01:16
jlgdr wrote: shobhitb wrote: Attachment: Odds.gif The chart below lists the odds that a horse will place in the top three. As part of a contest, if at least 2 of the horses that a person bets upon finish in the top three, the bettor receives a Tshirt. If Cecilia bets on The Baron, Happy Cynic, and Inamorata, what is the probability that she does not receive a Tshirt? A. 3/50 B. 21/100 C. 29/100 D. 35/100 E. 65/100 Im not sure wjere am I going wrong here. Would appreciate some guidance. We need to find the probability that either 2 or 3 finish on the top 3. So first for all of 'em to finish in the top three he have (3/5)(7/10)(1/2) Now for two of the three to finish top three we have (3C2)(3/5)(7/10)(1/2) This is not correct. Probability that two of the three will finish in the top three will be the sum of the individual cases: (3/5)(7/10)(1/2) + (3/5)(3/10)(1/2) + (2/5)(7/10)(1/2) Baron & Cynic + Baron & Inamorata + Cynic and Inamorata Note that the probability of each case is different since the probability of winning of each horse is different. Hence 3C2*(3/5)(7/10)(1/2) is not correct.
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Re: The chart below lists the odds that a horse will place in
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04 Mar 2014, 18:48
Bumping to this one. Still having trouble and getting D which is 35/100
Allow me to show my procedure:
1  (At least 2) 1  (2 or 3 horses finish at top spots)
Therefore
1  (3/5)(7/10)(1/2)+(3/5)(3/10)(1/2)+(2/5)(3/10)(1/2)+(3/5)(7/19)(1/2)
So we finally get 165/100= 35/100
I'm trying to figure out what is wrong here, any appreciation will be more than welcome
Thanks Cheers J



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Re: The chart below lists the odds that a horse will place in
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05 Mar 2014, 08:16
jlgdr wrote: Bumping to this one. Still having trouble and getting D which is 35/100
Allow me to show my procedure:
1  (At least 2) 1  (2 or 3 horses finish at top spots)
Therefore
1  (3/5)(7/10)(1/2)+(3/5)(3/10)(1/2)+(2/5)(3/10)(1/2)+(3/5)(7/19)(1/2)
So we finally get 165/100= 35/100
I'm trying to figure out what is wrong here, any appreciation will be more than welcome
Thanks Cheers J Man , this was a tricky question so i was stuck for a while but now i know what went wrong with your and my approach !!! I can help you !! There are 5 horses and there are 3 horses we are considering in our question , the probability that all three DO NOT FINISH in the top three should not be considered because out of 5 horses there has to be three finishers which means that at least one of the horses she picked will finish in the top three so 35/100(2/5*3/10*1/2) = 29/100 !! This was tricky but hope i was able to clear the issue for you



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Re: The chart below lists the odds that a horse will place in
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18 May 2014, 04:28
Got it right..Had to guess...As I got a 29 in my answer but had some other crap along with it too..So choose C (Took 3:47) Used the same logic in the correct answer except that I considered all 5 horses..i.e the probability that other two win.. 2/5*3/10*3/4*1/4*1/2 + 2/5*7/10*3/4*1/4*1/2 + 3/5*3/10*3/4*1/4*1/2 = 29*3/(50*32).. Had a option with 29 so Choose it.. Why are we not considering the other 2?
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The chart below lists the odds that a horse will place in
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31 May 2016, 03:36
I could get 29/100 but I have one query. Could the word "odds" in the question mean Odds in favour of? So the figures given will not be probabilities but odds in favour of particular horses winning. In that case, for example, the probability of Baron winning will be 3/8. Please suggest. chetan2u, Bunuel.



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The chart below lists the odds that a horse will place in
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31 May 2016, 04:11
shalinkotia wrote: I could get 29/100 but I have one query. Could the word "odds" in the question mean Odds in favour of? So the figures given will not be probabilities but odds in favour of particular horses winning. In that case, for example, the probability of Baron winning will be 3/8. Please suggest. chetan2u, Bunuel. Hi shalinkotia, I find the Q flawed on two accounts 1) The very point you have mentioned.Odds of winning = chances for:chances against.. Odds are not the same as probability 2) It is nowhere given that there are ONLY these 5 in race..so there is no reason why the prob that all three are not in TOP 3 is not to be taken.
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Re: The chart below lists the odds that a horse will place in
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31 May 2016, 11:39
Thank you chetan2u for the explanation.



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Re: The chart below lists the odds that a horse will place in
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16 Jul 2016, 03:05
I have a doubt. If I calculate the probability that she wins a tee, it'd be >3 cases wherein 2 horses win and one loses. >Or all three win.
(3/5)(7/10)(1/2) + (2/5)(7/10)(1/2) + (3/5)(3/10)(1/2) + (3/5)(7/10)(1/2)[All three win]=65/100 Probablity that she doesn't win=1Probablity that she wins Hence,35/100 Where am I going wrong?



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Re: The chart below lists the odds that a horse will place in
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22 Jul 2016, 09:55
shouldnt the answer be 3/16?
If California Girl and Love Song both win (3/4 * 1/4 = 3/16), it would be impossible for 2 other horses to win. Or am I missing something?



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Re: The chart below lists the odds that a horse will place in
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07 Nov 2017, 01:03
SlippinJimmy wrote: shouldnt the answer be 3/16?
If California Girl and Love Song both win (3/4 * 1/4 = 3/16), it would be impossible for 2 other horses to win. Or am I missing something? This is the same thing I thought I think the data are not coherent



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Re: The chart below lists the odds that a horse will place in
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03 Dec 2017, 22:58
Why aren't we considering the case where all three of them end up in the Top 3?
This was my solution: P(Tshirt) = (3/5 * 7/10* 1/2) + (3/5 * 1/2 * 3/10) + (1/2 * 7/10 * 1/2) + P(All three finishing in the Top 3) =(3/5 * 7/10* 1/2) + (3/5 * 1/2 * 3/10) + (1/2 * 7/10 * 1/2) + (3/5 * 7/10* 1/2) = 65/100
P(no Tshirt) = 35/100
Do we not consider all the horses in top three since the earlier three cases encompasses that? Someone help out please




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