Please tell me where I am going wrong.

A

Alpha6

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The Eagle's freshman football team is divided into 'Team A' and 'Team B'

70 percent are on team A and 30 percent are on team B.

Players on Team A have a 3 percent chance of injury, and players on Team B have a 2 percent chance of injury...

1. What is the probability a randomly chosen player will be on Team A and uninjured?

What I did was took the probability the player was on Team A... .7 and multiplied it by the chance they would not be injured. 1-.03 = .97 .... .97*.7 = .679


My conclusion was that a randomly chosen player had a .679 chance of being on team A and uninjured, I feel I messed up somewhere. Can anyone offer some guidance please?
 
Alpha6 said:
The Eagle's freshman football team is divided into 'Team A' and 'Team B'

70 percent are on team A and 30 percent are on team B.

Players on Team A have a 3 percent chance of injury, and players on Team B have a 2 percent chance of injury...

1. What is the probability a randomly chosen player will be on Team A and uninjured?

What I did was took the probability the player was on Team A... .7 and multiplied it by the chance they would not be injured. 1-.03 = .97 .... .97*.7 = .679
Click to expand...
That will work.
Let \(\displaystyle U\) be the event that the player is uninjured.
You are looking for \(\displaystyle \mathcal{P}(A\cap U)\).

Now \(\displaystyle \mathcal{P}(A\cap U)=\mathcal{P}(U|A) \mathcal{P}(A)\)
 
Last edited:
pka said:
That will work.
Let \(\displaystyle U\) be the event that the player is uninjured.
You are looking for \(\displaystyle \mathcal{P}(A\cap U)\).

Now \(\displaystyle \mathcal{P}(A\cap U)=\mathcal{P}(U|A) \mathcal{P}(A)\)
Click to expand...

Thanks for replying.

So P(A) = .7

and

P(U/A) = .97 ?
 
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