Ms. Barton has four children

[marc.gagnon]

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!

[jnelson0612]

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.

[marc.gagnon]

Thanks Jamie!

[jnelson0612]

My pleasure! :-)

[sukhpreetssekhon]

Hi Jamie,

I have a question here. Why does the order matter?
Can it not be divided as 3 cases.
GGBB - 1st case
GGGB - 2nd case
GGGG - 3rd case
Of course, I'm not close to the answer but it would be helpful if you could explain the logic.

Thanks

[RonPurewal]

In a probability scenario, you can only count "cases" if they are equally likely. Otherwise, the probability fraction becomes meaningless.
As a rather extreme example, consider playing the lottery. You can't say that the probability of winning the lottery is 1/2 because there are two "cases" (win and lose)! To correctly calculate the probability of winning the lottery, you must correctly break "win" down into all the possible ways of winning (= very few), and you must likewise break "lose" down into all the possible ways of losing (= lots and lots).

Consider your case "GGBB", vs. your case "GGGG".
There are six different ways in which a couple can have 2 girls and 2 boys (enumerated by Jamie above). By contrast, there's only one way to have four girls.
Therefore, it is six times as likely that a couple will have 2 children of each sex as that they will have four girls.

[RonPurewal]

If you're still having trouble with the boy/girl situation, you can always think about flipping 4 coins instead. (This example is more familiar to many students, because it's much more commonly presented in textbooks.)

If you flip 4 coins (or flip one coin 4 times in a row), there are six ways to get 2 heads and 2 tails:
hhtt
htht
htth
thht
thth
tthh
By contrast, there's only one way to get four tails: tttt. Same deal with the kids.