altairahmad wrote:
Hi
EMPOWERgmatRichC generis chetan2u VeritasKarishmaHow can I determine that permutations are required here ? Any thumb rule ?
The probability of getting 2 boys and 2 girls in any order is 1/16. By the wording of the question it appears that the question is asking for the probability of any one of these permutations of BBGG (as the end result will be same). Why are we considering BBGG to be different from BGBG ?
Please help to get a hang of this concept.
What are the total ways of having 4 children?
The first child in 2 ways, second in 2 ways, third in 2 ways and fourth in 2 ways to give 2^4 = 16 ways
This will include all cases: BBBB, BBBG, BBGB, BGBB, GBBB, BGBG, BBGG, GGBB, GBGB, GBBG, BGGB, GGGB, GGBG, GBGG, BGGG, GGGG etc
We need those which has 2 boys and 2 girls. How many such cases are there?
BGBG, BBGG, GGBB, GBGB, GBBG, BGGB (every way in which 2 boys and 2 girls are arranged is acceptable)
Since arrangement doesn't matter, we need to accept every arrangement (and that is the reason you need to find the number of arrangements)
Probability = 6/16 = Acceptable cases/Total number of cases
_________________
Karishma
Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >