SOLUTION: Each soccer player gets 4 jerseys red, yellow, white and green. A player randomly draws 2 jerseys each from the bag. The first jersey is not placed back in the bag. What is the

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Question 329387: Each soccer player gets 4 jerseys red, yellow, white and green. A player randomly draws 2 jerseys each from the bag. The first jersey is not placed back in the bag.
What is the probability of first drawing a white and then a green jersey?
What is the probability of drawing a red and yellow jersey in any order?
What is the probablity of not drawing a red jersey?
What is the probablity of having a white or red jersey after the second draw?
I would appreciate your help.
Thank you

Answer by Edwin McCravy(20086)   (Show Source): You can put this solution on YOUR website!
Each soccer player gets 4 jerseys red, yellow, white and green. A player randomly draws 2 jerseys each from the bag. The first jersey is not placed back in the bag.
A. What is the probability of first drawing a white and then a green jersey?

We (1) draw a white AND (2) then draw a green.

We start with this set of 16

R, R, R, R, Y, Y, Y, Y, W, W, W, W, G, G, G, G

(1) We can draw a white any of 4 ways out of 16.  That's a probability of ,  which reduces to ,

AND then that leaves this set of 15:

R, R, R, R, Y, Y, Y, Y, W, W, W, G, G, G, G

(2) We can then draw a green one any of 4 ways out of 15.  That's a probability of .

That's all we wanted to do.  Since the big word between 1 and 2 is AND, we multiply those probabilities:

Answer 







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B. What is the probability of drawing a red and yellow jersey in any order?

  What is the probability of first drawing a white and then a green jersey?

We (1) draw a red AND (2) then draw a yellow OR we (3) draw a yellow AND (4) then draw a red.

We start with this set of 16

R, R, R, R, Y, Y, Y, Y, W, W, W, W, G, G, G, G

(1) We can draw a red any of 4 ways out of 16.  That's a probability of ,  which reduces to ,

AND then that leaves this set of 15:

R, R, R, Y, Y, Y, Y, W, W, W, W, G, G, G, G

(2) We can then draw a yellow any of 4 ways out of 15.  That's a probability of .

OR

We start with this set of 16

R, R, R, R, Y, Y, Y, Y, W, W, W, W, G, G, G, G

(1) We can draw a yellow any of 4 ways out of 16.  That's a probability of ,  which reduces to ,

AND then that leaves this set of 15:

R, R, R, R, Y, Y, Y, W, W, W, W, G, G, G, G

(2) We can then draw a red one any of 4 ways out of 15.  That's a probability of .

 Red  AND yellow OR yellow  AND red   

AND means to multiply and OR means to add:





Answer 







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C. What is the probablity of not drawing a red jersey?

We (1) draw a non-red AND (2) then draw another non-red.

We start with this set of 16

R, R, R, R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R

(1) We can draw a non-red any of 12 ways out of 16.  That's a probability of ,  which reduces to ,

AND then that leaves this set of 15:

R, R, R, R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R, non-R 

(2) We can then draw a non-red any of 11 ways out of 15.  That's a probability of .

That's all we wanted to do.  Since the big word between 1 and 2 is AND, we multiply those probabilities:

Answer 







-----------------------------------

D. What is the probablity of having a white or red jersey after the second draw?

This requires finding the probability of the complenment event, and
then subtracting from 1

COMPLEMENT EVENT:

(1) Drawing a non-white-non-red AND (2) then drawing another non-white-non-red 

We start with this set of 16

R, R, R, R, W, W, W, W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W

(1) We can draw a non-red-non-white any of 8 ways out of 16.  That's a probability of ,  which reduces to ,

AND then that leaves this set of 15:

R, R, R, R, W, W, W, W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W, non-R-non-W

(2) We can draw a non-red-non-white any of 7 ways out of 15.  That's a probability of .

Answer to COMPLEMENT event:





Answer to DESIRED event:







Edwin

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