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
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
probability
This topic has expert replies
-
joshi.v123
- Junior | Next Rank: 30 Posts
- Posts: 19
- Joined: Wed Aug 12, 2009 9:34 pm
- Location: india
- Thanked: 3 times
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) ...
but i think that this question asks probability of getting one Black - white and one color set (or vice versa) ...
-
beat_gmat_09
- Master | Next Rank: 500 Posts
- Posts: 437
- Joined: Sat Nov 22, 2008 5:06 am
- Location: India
- Thanked: 50 times
- Followed by:1 members
- GMAT Score:580
two ways of getting -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
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
Hope is the dream of a man awake
-
Deepthi Subbu
- Master | Next Rank: 500 Posts
- Posts: 298
- Joined: Tue Feb 16, 2010 1:09 am
- Thanked: 2 times
- Followed by:1 members
-
Geva@EconomistGMAT
- GMAT Instructor
- Posts: 905
- Joined: Sun Sep 12, 2010 1:38 am
- Thanked: 378 times
- Followed by:123 members
- GMAT Score:760
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.
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.
