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A gumball machine contains 7 blue, 5 green, and 4 red [#permalink]
22 Oct 2012, 00:22

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A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only! _________________

If you find my post helpful, please GIVE ME SOME KUDOS!

Re: A gumball machine contains... [#permalink]
22 Oct 2012, 01:35

1

This post received KUDOS

Total No. of ways to select 3 out of 16 is = 16C3

No. Of ways to select one blue ball = 7C1 No. Of ways to select one green ball = 5C1 No. Of ways to select one red ball = 4C1

So. total no. of ways for favourable outcome = 7*5*4 = 140

So Probability = 140/560 = 1/4

When using simple probability, after choosing the first ball, there are only 15 balls to choose from and one less of the color that has already been chosen and after selecting the second, there are only 14 balls to choose from and one less of the color that has already been chosen and so on. Hence order has to be taken into account.

However while using combinations, the order does not matter as we are selecting one ball from one color only each time. _________________

Did you find this post helpful?... Please let me know through the Kudos button.

Re: A gumball machine contains... [#permalink]
22 Oct 2012, 02:49

jordanshl wrote:

I tried to solve using combinatorics but I was wrong - answer below.

I thought there is 16 balls in total and if we select three from 16, we have 16!/3!13! ways to choose three balls.

BGR is one of each and there is 6 ways of organising

we have 1C7x 1C5 x 1C4 x 6 / number of ways to choose three balls

so we have 7 x 5 x 4 x 6 / 16 x 15 x 14 x 3 x 2

Simplifying the sixes 7 x 5 x 4 / 16 x 15 x 14

simplifying 7 and 14 1 x 5 x 4 / 16 x 15 x 2

Simplifying 5 and 15 1 x 1 x 4 / 16 x 3 x 2

Simplifying 4, 16 1 x 1 x 1 / 4 x 3 x 2

Answer would be 1/24

However

Using simple probabilities

6x 7/16 x 5/15 x 4/14 = 1/4

You did a computational mistake, the 3! from the denominator of 16C3 should go up to the numerator. So your answer would be correct (1/24) x 6 =1/4. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: A gumball machine contains... [#permalink]
22 Oct 2012, 07:35

Expert's post

jordanshl wrote:

A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only!

We need to find the probability of BGR (a marble of each color).

Probability approach:

\(P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}\), we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).

Re: A gumball machine contains... [#permalink]
09 Jul 2013, 07:05

Bunuel wrote:

jordanshl wrote:

A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only!

We need to find the probability of BGR (a marble of each color).

Probability approach:

\(P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}\), we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).

Re: A gumball machine contains... [#permalink]
09 Jul 2013, 23:14

1

This post received KUDOS

Expert's post

elementbrdr wrote:

Bunuel wrote:

jordanshl wrote:

A gumball machine contains 7 blue, 5 green, and 4 red gumballs - identical besides colour. If the machine disperses 3 gumballs at random, what is the probability that it dispenses one of each colour.

Use the comibatorics method only!

We need to find the probability of BGR (a marble of each color).

Probability approach:

\(P(BGR)=\frac{7}{16}*\frac{5}{15}*\frac{4}{14}*3!=\frac{1}{4}\), we are multiplying by 3! since BGR scenario can occur in several different ways: BGR, BRG, RBG, ... (# of permutations of 3 distinct letters BGR is 3!).

Is there a way to solve this using \(\frac{16!}{7!*5!*4!}\) as the denominator?

Yes.

Total # of outcomes: \(\frac{16!}{7!*5!*4!}\) (the number of arrangements of the mables).

Favorable outcomes: we need the first three marbles to be BGR in any combination, so 3!. The remaining 13 marbles (6 blue, 4 green, and 3 red) can be arranged in 13!/(6!4!3!).

P = (Favorable)/(Total) = \(\frac{(3!\frac{13!}{6!4!3!})}{(\frac{16!}{7!*5!*4!})}=\frac{1}{4}\).

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