A gumball machine contains 7 blue, 5 green, and 4 red gumballs - ident

Question

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.

(A) 1/24 
(B) 1/4 
(C) 7/24 
(D) 1/3 
(E) 1/2

Use the comibatorics method only!

Bunuel · Accepted Answer

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!). Combination approach: P(BGR)=\frac{C^1_7*C^1_5*C^1_4}{C^3_{16}}=\frac{1}{4} . Hope it's clear. P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html (Please pay attention to the rule #8: Post Answer Choices for PS Questions).

Arthur1000 · Answer

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

MacFauz · Answer

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.

EvaJager · Answer

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.

elementbrdr · Answer

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

Bunuel · Answer

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} .

Hope it's clear.

altairahmad · Answer

Conceptual question please :

When we do 7 x 5 x 4 we are assuming that order matters and we will get BGR in every permutation. However, if we were to divide this by 3! what will we get ? (Yes its not a whole number)

MathRevolution · Answer

Blue = 7; Green = 5 ; Red = 4.

Total = 16.

'3' balls are dispersed =>  ^{16}\mathrm{C}_3

=>  ^{16}\mathrm{C}_3  =  \frac{16*15*14 }{ 3*2*1}  = 560 ways

The probability that it dispenses one of each color => Probability:  \frac{Desired }{ Total}

=>  ^7\mathrm{C}_1  = 7
 =>  ^5\mathrm{C}_1  = 5
 =>  ^4\mathrm{C}_1  = 4

=>  \frac{7 * 5 * 4 }{ 560}

=>  \frac{140}{560}

=>  \frac{1}{4}

Answer B

Kinshook · Answer

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

Favorable ways = 7*5*4 
Total ways = 16C3 = 16*15*14/3*2*1 = 16*5*7

Probability = 7*5*4/16*5*7 = 1/4

IMO B

Harshjha001 · Answer

I have one doubt . Can somebody guide me please.

In probability method i understand that (7/16∗5/15∗4/14∗3! ) , we are multiplying the result by 3! because the order is assumed as BGR . But in combination method why we don't multiply it by 3! ?

Thanks in advance.

Adarsh_24 · Answer

I’m very late to the party.

Anyway for others with similar doubts,

We actually do multiply.
7c1*5c1*4c1 / 16c3

The 3! from 16c3 will go to numerator.

Venu01298 · Answer

Any insights please on where it is appropriate to use the probability approach or combinations approach ? Just trying to know if there are any question types where one is preferred over the other

Thanks

Bunuel · Answer

In most cases, probability questions can be solved using multiple approaches. It often comes down to personal preference and the specific problem at hand when deciding which method to use. Some problems may feel more intuitive with a probability approach, while others might be easier to solve using combinations. Both methods are valid as long as they are applied correctly.

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A gumball machine contains 7 blue, 5 green, and 4 red gumballs - ident

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