QR Probability Question

[UCSD1984]

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. 16/33
B. 32/132
C. 12/32
D. 56/132
E. 32/144

This is how I approached it. 8red+4white = 12total. Probability of picking 1red = 8/12. Probability of picking 1white (i.e., not red) = 4/11 (because there's 4 white balls and 11 balls left total since 1 red one was already picked).

Then, I multiplied. (8/12) x (4/11) = 8/33. The answer, however, is A, 16/33. I'm stumped.

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[threechins]

You can either pick the red one first or second. So multiply 8/33 by 2 to get 16/33.

 

[jefff]

i do it in a similar way but really long :\

(red will bolded)

since its in draws of 2, make 2 lines
since its without replacement there will be 2 possible orders, make 2 sets of 2 lines

___ ___ ---> (8/12) (4/11) this is picking red first then white

(32)/(132) = (8*4)/(4*33) therefore 8/33


___ ___ ---> (4/12) (8/11) this is picking white first then red

(32)/(132) = (8*4)/(4*33) therefore 8/33



add the 2 possibilities together 16/33

 

[UCSD1984]

You can either pick the red one first or second. So multiply 8/33 by 2 to get 16/33.

Hmm, alright. Thanks.

 

[NewBee12]

If the order of picking the balls is not specified then you have to look at all the possibilities. If for example they want the probability of picking the red ball first then the white ball the answer will be 8/33.