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getzgetzu
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There are 8 TV sets, 6 are color and 2 are black and white. 26 Apr 2006, 03:38
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There are 8 TV sets, 6 are color and 2 are black and white. If 2 sets are selected at random, what is the probability that atleast one of them is color.



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klsrikanth
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There are 8 TV sets, 6 are color and 2 are black and white. 26 Apr 2006, 03:41
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(6c1 * 2c1) + 6c2 = 27
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getzgetzu
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There are 8 TV sets, 6 are color and 2 are black and white. 26 Apr 2006, 03:50
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Please explain all the answers you are posting. Stating a one line solution is not much of help!
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BG
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There are 8 TV sets, 6 are color and 2 are black and white. 26 Apr 2006, 03:55
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hallo
The prob that there is AT LEAST one color TV is equal to 1-prob there is none. Then total possibilities-2C8=28 , when only BW are selected-2C2=1 and the prob required is 1-1/28=27/28
hope it is of use
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klsrikanth
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There are 8 TV sets, 6 are color and 2 are black and white. 26 Apr 2006, 03:59
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selecting atleast a color set......

picking one color set out of 6 and one black setout of 2 = 6c1*2c1

picking two color sets from 6 = 6c2

(as arrangement is not a problem we will use combination here)

therefore combining two we get = (6c1*2c1) + (6c2)
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There are 8 TV sets, 6 are color and 2 are black and white. 26 Apr 2006, 04:02
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The fastest way is to find the probability that none is a color set and subtract that from 1.

P(none is color) = 2/8 * 1/7 = 2/56

P(at least one is color) = 1 - 2/56 = 54/56 = 27/28
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There are 8 TV sets, 6 are color and 2 are black and white. 28 Apr 2006, 23:14
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ywilfred
The fastest way is to find the probability that none is a color set and subtract that from 1.

P(none is color) = 2/8 * 1/7 = 2/56

P(at least one is color) = 1 - 2/56 = 54/56 = 27/28
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Agree with ywilfred's working



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