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A statistician reached the following conclusions about games between

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Akshaynandurkar

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19 Oct 2024, 02:51

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"When that team is the home team ..." is very important line to understand to answer this question correctly. "That team " here is referred to the team from the previous sentence - the team that scores the first goal . Now there is more than 50% chance the team that scores the first goal is likely to score that goal in first half and if THAT team is the home team then there is more than 50% chance that opponent of THAT team scores at least 1 goal

hence the answer is IF the home team scores the first goal then the opponent team scores one or more goals

parkhydel

A statistician reached the following conclusions about games between university soccer teams: Overall, a team playing on its home field has a 45% chance of a win, a 25% chance of a loss, and a 30% chance of a draw (a tied outcome). In the games where one or more goals are scored, the team that scores the first goal has a 55% chance of scoring it in the game's first half and a 45% chance of scoring it later in the game. When that team is the home team (i.e., a team playing on its home field), there is a 40% chance that the other team will score no goals at all, and therefore a 60% chance that it will score one or more goals.

Select for X and for Y two different outcomes such that the information provided explicitly includes the statistician's estimates of the probability that if X occurs, so will Y. Make only two selections, one in each column.

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kayarat600

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05 Dec 2024, 09:11

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OK, I spent 5 minutes on this but I figured out the reason this is in non math related and an official question it's just eliminating the options until you find one that has a follow up mentioned in the stem and choose it it's the fastest and least confusing method to solve this.

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12 Dec 2024, 04:31

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chetan2u

The probability scenarios are

Team playing on Home ground: W-45%, L-25% and D-30%
Games where at least one goal scored: First goal in first half - 55%; If home team scores first goal, 60% chances the other team will score at least one goal

Let us look at the question and the options:
So, we are looking at X happening translating in more than 50% chances of Y happening.
If X, then more than 50% chances of Y happening.

Options: Let us take each as X, and see if some other option fits in as Y.

A. The home team scores at least two goals in the game.
We do not have a scenario where home team scores at least two goals resulting in some other option.

B. The home team scores the first goal.
Look at the point (2) in brown. If 'The home team scores the first goal.' is happening, then 60% probability that other team will score at least one goal. Option E gives us the exact words, so would fit in as Y.

C. A goal is scored in the first half of the game.
If ' A goal is scored in first half(55%), then Y can be 'the games where one or more goals are scored'. But that has to be true and we do not have any option fitting in as Y.

D. A goal is scored in the second half of the game.
Is it the first goal of the match or one of many scored? No situation in the para, so does not fit in as X.

E. The team opposing the home team scores at least one goal.
There is nothing connected to this option, so cannot fit in as X. However it is connected and dependent on another situation given in option B, and can fit in as Y.

X: The home team scores the first goal.
Y: The team opposing the home team scores at least one goal.

Sorry, I don’t get the need for 50%.

Question asks for any explicit probability estimate for Y provided X happened.
I don’t see any other explicit probability for a final sequential event other than OA. Even if it was say 20% I think it would have fit.

Am I missing something?

avohra223

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Guys, I also struggled with this question for quite a while too. But I feel that students in this discussion are thinking about it wrongly. This is not a stats or a quant question.

It is truly just a verbal/RC style question, where you have to identify/recognise whether or not the STATISTICIAN'S ESTIMATE of the probabilities of a certain event to take place is provided in the above paragraph/the information provided to you.

Look closely at the other options. The information provided DOES NOT contain the probability estimates of those events, as given by the statistician. Take any other X,Y combination, and you will see that the information provided DOES NOT give us the statistician's estimate of the probabilities of that event (in that order) happening.

It's not a stats question saying "if X happens, then what WILL happen with 100% probability". It is just saying, "do we know the stats dude's estimate of the probability of this happening". And the inference "If X happens then what is the probability of Y happening by more than 50%" to me seems blatantly wrong. GMAC isn't a fool - and that is definitely a poor inference in my opinion.

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15 Jan 2025, 08:51

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This is an excellent way of looking at things!
Basically saying hey if you know X is it sufficient for you to find the exact probability from info given of Y happening!

avohra223

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rishabhj197

Here how do you decide that 'A goal is scored in the first half(55%) is X and 'the games where one or more goals are scored' is Y and not the other way round. Having a tough time establishing which is the bigger set here.

official explanation

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16 May 2025, 08:47

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I spent 10 minutes on this one and thought all answers are wrong. It turns out that I missed several words.
The question asks [explicitly includes the statistician's estimates of the probability that if X occurs, so will Y].
I just skimmed through the bold and underlined portion and thought it asks [if X happen, then Y must happen]. Didn't pay attention to the prompt, which asks only if I could deduce the Probability of Y...

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07 Jun 2025, 11:37

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Official Answer Explanation (from GMAC):

Recognize

The key information for determining outcomes X and Y such that if X occurs, then so does Y is “When that team (where ‘that team’ refers to the team that scores the first goal) is the home team ..., there is ... a 60% chance that it (where ‘it’ refers to the other team) will score one or more goals.” Therefore, from the given information, if the home team scores the first goal, the statistician’s estimate of the probability that the team opposing the home team scores at least one goal is 0.6.

RO1: Recognize

The correct answer is The home team scores the first goal.

RO2: Recognize

The correct answer is The team opposing the home team scores at least one goal.

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08 Jun 2025, 01:44

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I also got confused with the phrase "the probability that if X occurs, so will Y", and here's what Claude has to say about it:

Quote:

The statement "the probability that if X occurs, so will Y" refers to the likelihood of Y happening given that X has already occurred. In other words, it asks: "If X happens, how likely is it that Y will also happen?", i.e. conditional probability - p(Y|X)

And that resolves my doubt, and the answer seems in coherence with this logic.

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27 Sep 2025, 01:02

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Honestly I hate this question, but here is what I finally was able to understand:

Events:

W -> Team wins
L -> Team loses
D -> Team Draws
H -> Home Team
F -> Team that scores the first goal
H1 -> First goal in First half
H2 -> First goal in second half
OP1 -> Opponent scores at least one goal
OP0 -> Opponent scores 0 goals

What is the question asking, find for X and Y such that the statisticians estimate for P(Y|X) is given.

So let's put the statisticians estimates into the form P(X|Y):

P(W|H) = 0.45 (Given that the team is home team, the prob of that team winning)
P(L|H) = 0.25 (Given that the team is home team, the prob of that team losing)
P(D|H) = 0.3 (Given that the team is home team, the prob of that team drawing)

P(H1|F) = 0.55 (Given that the team is the one that scores the first goal, the prob of that goal happening in the first half)
P(H2|F) = 0.45 (Given that the team is the one that scores the first goal, the prob of that goal happening in the Second half)

P(OP1| F & H) = 0.6 (Given that the team that scores the first goal is the home team, the prob that the opponent scores at least one goal)
P(OP0| F & H) = 0.4 (Given that the team that scores the first goal is the home team, the prob that the opponent scores 0 goals)

Now let's look at the options:

The home team scores at least two goals in the game. (Scoring multiple goals is talked about for the opponent not the home team. This prob is not given)
The home team scores the first goal. (We have probabilities where this happens P(OP1| F & H) & P(OP0| F & H))
A goal is scored in the first half of the game. (We don't care about a goal, but instead the first goal. None of the prob given are related to just a goal)
A goal is scored in the second half of the game. (We don't care about a goal, but instead the first goal. None of the prob given are related to just a goal)
The team opposing the home team scores at least one goal. (We have a probability where this happens P(OP1| F & H))

So we have exactly P(OP1 | F & H) = 0.6. So therefore X = The home team scores the first goal. (F&H) ; Y = The team opposing the home team scores at least one goal. (OP1)

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