Marble color probability

Probability Level 3

A bag contains 4 of each color of marble: red, orange, green, blue, and purple. 3 marbles are drawn from the bag (without replacement). If $\frac{a}{b}$ is the probability that the drawn marbles are different colors, where $a$ and $b$ are coprime positive integers, then what is $a+b?$

The answer is 89.

2 solutions

Andy Hayes
Dec 14, 2016

Treat each marble in the bag as distinct. Consider the combinations of 3 marbles that are different colors. There are $\binom{5}{3}$ ways to select the colors of those 3 marbles. Then, there are $\binom{4}{1}^3$ ways to select the marbles from each of those colors. The total number of ways to select 3 marbles from the bag is $\binom{20}{3}.$ Thus, the probability that the 3 drawn marbles are different colors is:

$\frac{\binom{5}{3}\binom{4}{1}^3}{\binom{20}{3}}=\frac{640}{1140}=\frac{32}{57}$

Thus, $a+b=\boxed{89}.$

U have not mentioned that each marble is distinct in the question.Due to this I am not able provide correct answer.

Kushal Bose - 4 years, 6 months ago

It isn't a necessary assumption. The marbles can be treated as distinct without changing the problem. Imagine taking the marbles of each color and labeling them with the numbers 1 through 4 to make them "distinct." The problem is the exact same, except now combinations can be used to compute distributions.

Identical marble distributions are somewhat more challenging to compute. There are $\binom{5}{3}+2\binom{5}{2}+\binom{5}{1}=35$ distributions of identical marbles. An issue with treating marbles as identical in the calculation is that these distributions are not uniform. For example, the distribution $RGB$ is more likely to occur than the distribution $RRG.$

Here is an alternative calculation using permutations of draws instead of combinations:

On the first draw, there are 20 marbles in the bag, and it doesn't matter which marble is chosen. On the next draw, there are 19 marbles left in the bag, and 16 of them are a different color than the first marble. If the first two marbles were different colors, then on the next draw, there are 18 marbles left in the bag, and 12 of them are a different color than the two drawn marbles. The probability that the three marbles are a different color is:

$\frac{20 \times 16 \times 12}{20 \times 19 \times 18}=\frac{3840}{6840}=\frac{32}{57}.$

Andy Hayes - 4 years, 6 months ago

why is there (4 choose 1)^3 ways to select the marbles from each of those colors?

Connie Chen - 2 years, 8 months ago

There are ( 5 choose 3 ) ways to select the colors of 3 marbres so there are for each 3 chosen marbres ( 4 choose 1) ways to select one. There are in total (4 choose 1)^3 ways to select 3 marbres from each of those colors.

Tedi Kakatsi - 2 years, 7 months ago

Laura Gao
Mar 29, 2018

The total number of combinations of marbles is $20*19*18$ . The possible arrangements where all the marbles are different colors are $20*16*12$ . The probability is $\(\frac{20*16*12}{20*19*18}$ = $\frac{32}{57}$ ). The answer is $32+57=89$ .

The answer is 89.

Marble color probability

The answer is 89.

2 solutions

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