The chart below lists the odds that a horse will place in

"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 T-shirt. If Cecilia bets on The Baron, Happy Cynic, and Inamorata, what is the probability that she does not receive a T-shirt?"

Cecilia can win a T-Shirt, if at least 2 of The Baron, Happy Cynic, and Inamorata finish in top three.

Now theoretically, when she will not get a T-Shirt ? 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 T-Shirt = 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) * (1-7/10)*(1-1/2) + (1-3/5) * (7/10)*(1-1/2) + (1-3/5) * (1-7/10)*(1/2)

= 29/100

The probability that it will not win the T-shirt=
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

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

#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?


_________________

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 t-shirt
and By 1- P(winning) = Not winning.

Is this procedure accurate?
Thanks

Cheers!
Kudos rain
J

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.

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 1-65/100= 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

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?

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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Thank you chetan2u for the explanation.

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=1-Probablity that she wins
Hence,35/100
Where am I going wrong?

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

Why aren't we considering the case where all three of them end up in the Top 3?

This was my solution:
P(T-shirt) = (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 T-shirt) = 35/100

Do we not consider all the horses in top three since the earlier three cases encompasses that? Someone help out please