Permutation Question (Please help)

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Moonlightbeam

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Jun 19, 2021
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I am totally lost on how to solve this multiple choice question. I don't know how to start it. Is there supposed to be an equation I am supposed to use.

United have 11 players on the pitch at the end of extra time, including De Gea and Slabhead. If the penalty kick order is randomly assigned, what is the probability that Slabhead will kick first, followed by De Gea? Note: De Gea is one footballer, Slabhead is another, these are two different players.
  • 2/121
  • 2/11
  • 1/121
  • 1/110
 
Moonlightbeam said:
I am totally lost on how to solve this multiple choice question. I don't know how to start it. Is there supposed to be an equation I am supposed to use.

United have 11 players on the pitch at the end of extra time, including De Gea and Slabhead. If the penalty kick order is randomly assigned, what is the probability that Slabhead will kick first, followed by De Gea? Note: De Gea is one footballer, Slabhead is another, these are two different players.
  • 2/121
  • 2/11
  • 1/121
  • 1/110
Click to expand...
If \(\mathcal{D}\) stands for DeGea & \(\mathcal{S}\) stands for Slabhead do you \(\mathcal{S}\) is the first player the kick and then \(\mathcal{D}\) is second player to kick?
Or do you mean that \(\mathcal{S}\) simply kicks before \(\mathcal{D}\) kicks. Or something else.
If it is that \(\mathcal{S}~\&~\mathcal{D}\) are first & second to kick, then that is \(\dfrac{9!}{11!}=\dfrac{1}{110}\)
 
pka said:
If \(\mathcal{D}\) stands for DeGea & \(\mathcal{S}\) stands for Slabhead do you \(\mathcal{S}\) is the first player the kick and then \(\mathcal{D}\) is second player to kick?
Or do you mean that \(\mathcal{S}\) simply kicks before \(\mathcal{D}\) kicks. Or something else.
If it is that \(\mathcal{S}~\&~\mathcal{D}\) are first & second to kick, then that is \(\dfrac{9!}{11!}=\dfrac{1}{110}\)
Click to expand...
The question is asking what is the probability that Slabhead will kick first, followed by De Gea if the penalty kick order is randomly assigned.
 
The question then is: what is the probability that S and D are in positions 1 and 2 respectively, for taking penalties?
You have [MATH]\tfrac{1}{11}[/MATH] chance of picking S first, then [MATH]\tfrac{1}{10}[/MATH] chance of picking D second. Prob =[MATH]\tfrac{1}{11}[/MATH] [MATH]\times[/MATH] [MATH]\tfrac{1}{10}[/MATH] = [MATH]\tfrac{1}{110}[/MATH]
 
If a player can take 2 penalty kicks (I'm no football expert) then the prob that S will be picked first is \(\displaystyle \frac{1}{11}\), then the prob that D will be picked next is also \(\displaystyle \frac{1}{11}\), so the prob that S then D kick in that order is \(\displaystyle \frac{1}{11}*\frac{1}{11} = \frac{1}{121}\).
UNLESS, the first kicker can't be chosen twice. Then the answers above ie \(\displaystyle \frac{1}{110}\) are correct.
 
@Harry_the_cat
Good point, but they initially only take one!


No player will be allowed to take a second kick until all other eligible players on their team have taken a first kick, including the goalkeeper.

If it becomes necessary for players to take another kick (because the score has remained equal after all eligible players have taken their first kick), players are not required to kick in the same order.
 
lex said:
@Harry_the_cat
Good point, but they initially only take one!


No player will be allowed to take a second kick until all other eligible players on their team have taken a first kick, including the goalkeeper.

If it becomes necessary for players to take another kick (because the score has remained equal after all eligible players have taken their first kick), players are not required to kick in the same order.
Click to expand...
Thanks Lex. So 1/110 is the correct answer. It's a very biased question against someone who doesn't know the rules though.
 
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