M14-29

Total goals = 5
Winner goals = 3
Loser goals = 2

P (Loser hit the first goal) = P (Winner did not hit the first goal) = 1 - P (Winner hit the first goal)
=> 1- 3/5
=> 2/5
C

The original question asks for the probability that the losing team scored the first goal in a match that ended 3:2. While it may be tempting to think that the probability is 50% since there are only two possibilities for the first goal, this line of thinking is not correct. The actual score of the match determines the probability of each team scoring a goal, and therefore the probability that the losing team scored the first goal.

Consider the scenario where the match ends with a score of 5:0. Would you still say that the probability of the losing team scoring first is 50% in this scenario? In this case, the probability that the losing team scored the first goal is 0%, not 50%. This is because the actual score of the match eliminates the possibility of the losing team scoring any goals. Similarly, in the original scenario where the match ends 3:2, the actual score of the match determines the probability of each team scoring a goal, and thus the probability that the losing team scored the first goal is not simply 50%.

Official Solution:

If a soccer match ended with a score of 3:2 and assuming that all possible scoring scenarios of the match had an equal probability, what is the probability that the team which ultimately lost had scored the first goal?

A. \(\frac{1}{4}\)
B. \(\frac{3}{10}\)
C. \(\frac{2}{5}\)
D. \(\frac{5}{12}\)
E. \(\frac{1}{2}\)


Let's start by determining the number of possible scoring scenarios for the match. We can represent a goal scored by the winning team with W and a goal scored by the losing team with L. The total number of possible scoring scenarios is the number of ways to arrange the letters (L, L, W, W, W), which is equal to \(\frac{5!}{2!3!}=10\).

To find the number of scenarios in which the losing team scored the first goal, we can fix L in the first position and arrange the remaining four letters (L, W, W, W) in \(\frac{4!}{3!}=4\) ways. Therefore, the probability that the team which ultimately lost had scored the first goal is \(P=\frac{favorable}{total}=\frac{4}{10}=\frac{2}{5}\).

Alternatively, we can consider the distribution of (L, L, W, W, W) into a grid representing the sequence of goals (first goal, second goal, third goal, fourth goal, fifth goal). What is the probability that we select an L when choosing which letter to put in the first goal position? Since there are two L's out of a total of five letters, the probability is \(P=\frac{2}{5}\).


Answer: C

I think this is a high-quality question and I agree with explanation.

Wow! Lesson learned. I was so set on using combinations to solve this that I completely missed the obvious and easy approach. Took me 3 mins to ensure that my probability was correct when I should have setup the answer the way you did and be done within 30 seconds.

Is the reason why this is not 1/2 because the potential outcomes have already been dictated by the score? In other words, if the question stem had not provided the score of the game and the question was "what is the probability that Team A scores first?", then that probability would be 1/2?

However, because you are given the final score, there are 10 possible ways the game occurred:

LLWWW
LWLWW
LWWLW
LWWWL
WWWLL
WWLWL
WWLWL
WWLLW
WLWWL
WLLWW
WLWLW

The first four outcomes are the only ways the team could have not gotten the first goal out of the ten outcomes. P = 4/10 = 2/5

Correct?

P.S. - Can someone please express this in a combinatoric approach? I understand that there are 5c2 ways to arrange LLWWW, but how do we get the numerator to be 4? I believe it's (2c1)*2, since order of "which" L doesn't matter, but can someone please confirm?

Yes, that's correct.

Although I got the question correctly, something that is still kind of a doubt is that, if we are asked the probability of the loosing team scoring first, and before the first goal is scored, wont the probability be 1/2 at the time of scoring the goal ? so idk if that makes sense but that was something that came to my mind while solving the problem but that seemed too far fetched and complicated so i choose the analog way of thinking about the question and went with 2/5

That's a good point. I like your thinking - at that point the probability of scoring is really 50/50 if the events are indeed independent but I think we don't know that both teams are equal in capability and this is not necessarily like flipping a coin that has a 50/50 chance. Any betting place will not give you a 50/50 chances for 2 teams ... I don't have an answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.