renatakater wrote:I didn't understand why the first girl is 1. In my opinion it should be 1/2, or 5/10.

Negin wrote:

Brent@GMATPrepNow wrote:

[email protected] wrote:Hi Brent can you please tell me that why did you take "1 and 1 later on in the following equation:

1*(4/9)*(3/8)*(2/7)*(1/6)=1/126

According to me it should be (1/10 * 4/9 * 3/8 * 2/7 * 1/6)
Please tell me why the 1 in the beginning wherein it should be (1/10) right as the girl will pick any ball initially.

Hi Amit,

You're missing something from my calculations.
It should be 1 x 4/9 x 3/8 x 2/7 x 1/6 x 1

The first 1 is the probability that the 1st girl selects any marble
The probability that she selects any marble is 1. In other words, we can guarantee that she will definitely select any marble.

The last 1 is the probability that the boys get same color marble from remaining marbles.
In our calculations, we are assuming that all of the girls choose the same colored marble. So, once the girls have selected their marbles, the 5 remaining marbles must be the same color.
If the 5 remaining marbles are the same color, the boys are guaranteed to all get the same color.
So, the probability that the boys get same colored marble = 1

I hope that helps.

Cheers,
Brent

Thanks For Brent solution, But we can change the solution a little which might be more clear:

If the first girl choose white marble then the probability of same marble color for all girls is 5/10*4/9*3/8*2/7*1/6
If the first girl choose black marble then the probability of same marble color for all girls is same as above. so
probability of same marble color for all girls=5/10*4/9*3/8*2/7*1/6*2=1/126

nischo wrote:Hi Brent, I have a quick qn here, Where in this solution are we considering the positioning of the girls or the boys, by positioning I mean the order of their picking.

kunalmoudgil wrote:*Brent - Can you offer any guidance on this? Much appreciate*

Bc, I didnt know the order of picking (whether all girls were picking before the boys, or whether they were picking alternatively etc.) I assumed that they would pick alternatively. So it was not only important that each girl pick the same colored marble, but also that the boys don't pick those marbles on their turn. So to me the probability would go something like this: 1 X P(boy not picking that marble) X P(girl getting that marble)...

The formula would look something like this:

1 X (1-4/9) X (4/8) X (1-3/7) X (3/6) X (1-2/5) X (2/4) X (1-1/3) X (1/2)
1 X (5/9) X (4/8) X (4/7) X (3/6) X (3/5) X (2/4) X (2/3) X (1/2)
5X4!X4! / 9! = 1/126

What do you think?! Long way to solve an easy question

nischo wrote:

Brent@GMATPrepNow wrote:

nischo wrote:Hi Brent, I have a quick qn here, Where in this solution are we considering the positioning of the girls or the boys, by positioning I mean the order of their picking.

Good question, nischo.

For this solution, the order of the picking doesn't matter.
I chose to have the girls pick first, but the solution will be the same no matter what order you use.

To illustrate this, consider the following rudimentary question: A box contains a white chip and a black chip. If you and I both select a chip at random, what is the probability that you select the black chip? The probability will be 1/2 regardless of the order in which we select chips.

Cheers,
Brent

Thanks for the quick reply.
But the order would matter if the distribution was unequal right. For example if in the same qn there were 6 girls and 4 boys(also 6 white and 6 black balls). I kno it makes the qn whole lot complicated, but I think this would help everyone understand the concept really well.