shavarna wrote:
I thought that order does not matter for when child is born, therefore
Total number of events = 5 (explained below)
0 Boys 4 Girls
1 Boy 3 Girls
2 Boys 2 Girls
3 Boys 1 Girl
4 Boys 0 Girl
Probability (2 B and 2 G exactly) = 1/5
Not sure where I am assuming and making a mistake.
Hi shavarna,
The 5 options you've listed are NOT all equally likely, so you would have to do a bit more work to get to the correct answer. There are a couple of ways to answer this question:
1) You could make a table of all the options (since each child could be a boy or a girl, there are only 2^4 = 16 possible outcomes) and determine all the ways to get 2 boys and 2 girls
or
2) You can do the math
Here's the math approach:
Since each child has an equal chance of ending up as a boy or girl, there are 2^4 possibilities = 16 possibilities. It also doesn't matter which 2 of the 4 children are boys, so you can treat this part of the question as a combination formula question...
4c2 = 4!/[2!2!] = 6 ways to get 2 boys and 2 girls out of 16 possibilities. 6/16 = 3/8
Final Answer =
GMAT assassins aren't born, they're made,
Rich
But that's also valid method right as nowhere stated in question that in how many different way this is possible.