Girls & Boys
MGMAT Challenge Problem 06/02/03

Question Statement
Ms. Barton has four children. You are told correctly that she has at least two girls but you are not told which two of her four children are those girls. What is the probability that she also has two boys? (Assume that the probability of having a boy is the same as the probability of having a girl.)

(A) 1/4
(B) 3/8
(C) 5/11
(D) 1/2
(E) 6/11

Solution
Since each of the 4 children can be either a boy or a girl, there are 2*2*2*2 = 2^4 = 16 possible ways that the children might be born, as listed below:

BBBB (all boys)
BBBG, BBGB, BGBB, GBBB, (3 boys, 1 girl)
BBGG, BGGB, BGBG, GGBB, GBBG, GBGB (2 boys, 2 girls)
GGGB, GGBG, GBGG, BGGG (3 girls, 1 boy)
GGGG (all girls)

Since we are told that there are at least 2 girls, we can eliminate 5 possibilities--the one possibility in which all of the children are boys (the first row) and the four possibilities in which only one of the children is a girl (the second row).

That leaves 11 possibilities (the third, fourth, and fifth row) of which only 6 are comprised of two boys and two girls (the third row). Thus, the probability that Ms. Barton also has 2 boys is 6/11 and the correct answer is E.

My Question
I understand how we arrive at 16 possible ways. Please help me understand how I can solve the second half of the problem (the numerator) other than writing out each of the various combinations.

I have tried various ways to use combinatorics, but I am not making any progress.

Thanks!

Okay, again, here are the possibilities:

BBBB (all boys)
BBBG, BBGB, BGBB, GBBB, (3 boys, 1 girl)
BBGG, BGGB, BGBG, GGBB, GBBG, GBGB (2 boys, 2 girls)
GGGB, GGBG, GBGG, BGGG (3 girls, 1 boy)
GGGG (all girls)

You could use the combinatorics/order matters formula to get each of these combinations.

If you are taking our class, you may remember that one of the class problems is to determine how many different arrangements there are for the word RADAR. We find it through this formula:
Pool!
Repeat! Repeat! Repeat!

In this case:
5!
2!2!1!

We write it this way because we have 5 letters, with two repeating Rs, two repeating As, and 1 D.

Using this formula, you could go through every combo of boys and girls and determine how many arrangements there are of Bs and Gs.

For example, 2 Bs and 2Gs would be:
4!
2! 2!
which results in 6. Note that this matches what we have above.

Also, 3 Bs and 1G (and 3Gs and 1B) would be:
4!
3! 1!
which results in 4. Again, note the match.

So you'd say 1 arrangement of all boys, 1 arrangement of all girls, 6 arrangements of 2 and 2, and 4 arrangements of 3B1G and 4 arrangements of 3G1B. 1+1+6+4+4 = 16.

_________________
Jamie Nelson
ManhattanGMAT Instructor

Thanks Jamie!

My pleasure! :-)

_________________
Jamie Nelson
ManhattanGMAT Instructor