Probability question

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299678

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A container holds 8 red and 4 white balls. Two balls are drawn in sequence without replacement from the container. Which of the following represents the probability that exactly one of these balls is red?
 
A container holds 8 red and 4 white balls. Two balls are drawn in sequence without replacement from the container. Which of the following represents the probability that exactly one of these balls is red?

if first ball is red, (8/12)*(4/11)

if second ball is read, (4/12)*(8/11)

add them up you got 6/11?
 
there is 1/2 chance for 1st ball being red, and same 1/2 chance for 2nd ball being red.
So instead of simply adding up (8/12)*(4/11) + (4/12)*(8/11), you should do (1/2)(8/12)*(4/11)+ (1/2)(4/12)*(8/11) = 32/132
 
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excuse me if this is a wrong answer, but can anyone tell me if this is right/wrong?

add up 8 red and 4 white = 12 total

Draw 1 (red): 8/12
Draw 2 (red): 7/11

P(red) = (8/12)*(7/11) = 56/132 - answer D
 
The guy way up there got it right (just added them incorrectly). No need to keep chiming in with wrong answers. The correct answer is A.

there is 1/2 chance for 1st ball being red, and same 1/2 chance for 2nd ball being red.
So instead of simply adding up (8/12)*(4/11) + (4/12)*(8/11), you should do (1/2)(8/12)*(4/11)+ (1/2)(4/12)*(8/11) = 32/132

Wrong. There is not a 1/2 chance you'll draw a red ball when there are clearly DOUBLE the amount of red balls to start.

excuse me if this is a wrong answer, but can anyone tell me if this is right/wrong?

add up 8 red and 4 white = 12 total

Draw 1 (red): 8/12
Draw 2 (red): 7/11

P(red) = (8/12)*(7/11) = 56/132 - answer D

Wrong answer. The question asked for 1 of each. You solved for 2 reds.
 
you have to calculate for both balls then add together,
8/12 * 4/11 = 8/33 next 4/12 * 8/11 = 8/33 now add together you get 16/33 answer A.