Problem Solving

If a certain soccer game ended 3:2, what is the probability that the side that lost scored first?
(Assume that all scoring scenarios are equiprobable)

1/4

3/10

2/5

5/12

1/2

OA: C

Would love to know where logic in choosing E went wrong. After all, [spoiler]first goal is like a coin toss, any team could have scored it equally, so it's a 50/50 chance.[/spoiler]

adilka wrote:If a certain soccer game ended 3:2, what is the probability that the side that lost scored first?
(Assume that all scoring scenarios are equiprobable)

1/4

3/10

2/5

5/12

1/2

OA: C

Would love to know where logic in choosing E went wrong. After all, [spoiler]first goal is like a coin toss, any team could have scored it equally, so it's a 50/50 chance.[/spoiler]

What a great question!!!

First of all, the flaw in your logic is since we know the score we can have N different scanorios to end up at 3:2 and since 3 is not equal to 2 there will more chances for winning team to score the first goal.

Actually it will be 2/(2+3) for the losing team and 3/(2+3) for the winning team!

Thanks Logitech. I do understand. But that 50/50 coin toss still bothers me Can't see how I could have avoided it.

The question here is which team is more likely to have scored first.

If Team A scored 3 times and Team B scored 2 times, you would have the following set of 5 goals:

A A A B B

From the group of five letters above, if you randomly pick one letter, what is the probability of picking A?

When you think of the first goal as a 50% chance for both teams, you are isolating the first goal and making it independent of the total number of goals scored. To make it more simpler and intuitive think of it this way: If team A scored all 5 goals, what is the probability that Team A scored the first goal. Obviously 100% right? That should help you see why it would not be 50%.

arzanr wrote:The question here is which team is more likely to have scored first.

If Team A scored 3 times and Team B scored 2 times, you would have the following set of 5 goals:

A A A B B

From the group of five letters above, if you randomly pick one letter, what is the probability of picking A?

When you think of the first goal as a 50% chance for both teams, you are isolating the first goal and making it independent of the total number of goals scored. To make it more simpler and intuitive think of it this way: If team A scored all 5 goals, what is the probability that Team A scored the first goal. Obviously 100% right? That should help you see why it would not be 50%.

Great explanation!

arzanr wrote:If team A scored all 5 goals, what is the probability that Team A scored the first goal. Obviously 100% right? That should help you see why it would not be 50%.

Great explanation indeed!

My solution is:
I have 5 goal three of one team (winner) two of another team (loser).
I have 5!/(2!*(5-2)!) = 10
possibilities of arrange a subgroup of two goal ( the one of the loser).
I want that one of the first goal of the loser is in the first place, so I have
4!/(3!*(4-3)!)=4
possible arrangements for the other goals.
Result is 4/10 probability that the first goal is of the loser ===> 2/5

adilka wrote:

arzanr wrote:If team A scored all 5 goals, what is the probability that Team A scored the first goal. Obviously 100% right? That should help you see why it would not be 50%.

Great explanation indeed!