Probability

[c.w.richardjr]

This question (not exact) comes from GMAT Prep.

From a group of 3 boys and 3 girls, 4 are selected randomly. What is the probability that an equal number of boys and girls will be selected?

Please help.

Charles

[vikash.121186]

I think it should be (3C2*3C2)/6C4 which is 9/15=3/5.

[jnelson0612]

This question (not exact) comes from GMAT Prep.

From a group of 3 boys and 3 girls, 4 are selected randomly. What is the probability that an equal number of boys and girls will be selected?

Please help.

Charles

Would you like to post the exact question? GMAT Prep questions are allowed to be posted here.

[c.w.richardjr]

I think it should be (3C2*3C2)/6C4 which is 9/15=3/5.

Thank you for your response. Could you possibly provide an explanation? First off, I am unfamiliar with the nCr formula, which I have now learned. Still, I was unaware of this being a tool to use for Combinatorics / Probability. How do you know when to use this tool? Please explain a little bit. Thank you.

Sorry but I don't have the full problem. It was off of the GMAT Prep practice test, and I didn't write the whole thing down.

[RonPurewal]

This question (not exact) comes from GMAT Prep.

From a group of 3 boys and 3 girls, 4 are selected randomly. What is the probability that an equal number of boys and girls will be selected?

Please help.

Charles

if you don't like the c/p formulas, you can always do the low-tech solution: just list out all the possibilities, and circle the ones that do what you want.
let's say the boys are a, b, c, and the girls are x, y, z. (better than d/e/f, because the difference is more obvious if you pick letters from opposite extremes of the alphabet.)
here are all ways to select 4 of them -- as it turns out, there are only 15 ways:
a, b, c, x
a, b, c, y
a, b, c, z
a, b, x, y
a, b, x, z
a, b, y, z
a, c, x, y
a, c, x, z
a, c, y, z
a, x, y, z
b, c, x, y
b, c, x, z
b, c, y, z
b, x, y, z
c, x, y, z

you should be able to make this list in substantially less than 1 minute.

now, just circle (= boldface, here) the ones that have exactly two of the boys (a, b, c) and exactly two of the girls (d, e, f):
a, b, c, x
a, b, c, y
a, b, c, z
a, b, x, y
a, b, x, z
a, b, y, z
a, c, x, y
a, c, x, z
a, c, y, z
a, x, y, z
b, c, x, y
b, c, x, z
b, c, y, z
b, x, y, z
c, x, y, z

that's a total of 9 out of 15 possibilities, giving a probability of 9/15 = 3/5.

[zishj865]

Hi Ron

I solved it with the probability approach. I know the combinations approach is much easier, but i want to understand the probability approach that dose not involve listing all the outcomes.

3/6. 2/5. 3/4.2/3 * 4!/2!2! =3/5

Here the order within the 2-boy and 2-girl groups has already been accounted for .When we multiply by 4! it will duplicate groups of boys and girls which have already been considered in the 3/6 *2/5* 3/4 *2/3 part.

e.g B1B2G1G2 would be repeated when we multiply by 4! because B1B2G1G2 has already been counted once in the 3/6* 2/5* 3/4 *2/3 so to eliminate those repeated combinations we need to divide by 2! twice.

Is my reasoning correct?


Thanks

[RonPurewal]

that reasoning looks valid to me... but, honestly,

...1/
how long did it take you to come up with that?

...2/
how CERTAIN were you of the result?
(as opposed to just listing out 15 possibilities and counting 9 of them, which would have absolute certainty)

AND
...3/
did you actually come up with that reasoning on your first try? or did you only get there because you knew the answer was 3/5, and so you could keep adjusting the reasoning until you actually got that answer?

remember—you should be trying for approaches that are actually practical under testing conditions.
the reasoning you've posted here is ... impressive, but, i don't think it has a lot of utility as far as actually preparing you for the test you're going to take.

in fact, if you would actually NOT just make a list of the 15 possibilities BECAUSE you were trying to come up with something "advanced" like this, then, thinking about these kinds of "advanced" approaches will actually make you WORSE at this test!