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by joshi.v123 » Sat Nov 13, 2010 5:29 am
is it 12/28?
Thanks,
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by mariah » Sat Nov 13, 2010 6:08 am
13/28 would be the answer if question asked what would b the probability if at least one black - white stem is selected...

but i think that this question asks probability of getting one Black - white and one color set (or vice versa) ...
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by beat_gmat_09 » Sat Nov 13, 2010 6:50 am
mariah wrote:we have six color TV and two black-white TV. If two sets are to be selected at random, what is the probability that one black-white set is selected?



sorry don't have answers and oa
two ways of getting -
BW C or C BW.
prob BW C = 2/8*6/7
prob C BW = 6/8*2/7
req prob - 2/8*6/7 + 6/8*2/7 = 3/7
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by ov25 » Sat Nov 13, 2010 7:25 am
1-30/56 = 26/56 = 13/28

thanks
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by Deepthi Subbu » Tue Nov 16, 2010 9:12 pm
Am unable to understand , can somebody explain it further?
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by Geva@EconomistGMAT » Wed Nov 17, 2010 2:56 am
Beat_gmat_09 is correct.

Break the problem down to 1st choice, 2nd choice, find the probability of success in each, then multiply.

It seems as if the answer is P(BW on 1st) * P(color on 2nd)) = 2/8 * 6/7 = 1/2 * 3/7 = 3/14

However, that is only one of two possible scenarios of getting "one B/W", the other scenario being P(color on 1st) * P(B/W on 2nd) = 6/8 * 2/7 = 3/4 * 2/7 = 3/2 * 1/7 = 3/14. Not surprising the probability for both scenarios comes down to the same: there really is no reason why the probability of choosing a B/W first should be any different from the probability of choosing a B/W 2nd.

Since the final probability is made of P(choosing B/W on 1st) OR P(choosing B/W on 2nd), the final answer is 3/14+3/14 = 6/14 = 3/7.
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