Why is the answer not 1/3?

Choosing a color uniformly would give 1/3, but the colors are not equally likely and the second marble is drawn from a smaller bag. Counting pairs gives 19 favorable out of 66, which is 19/66, not 22/66.

What changes if the first marble is put back?

Then the draws are independent: (5/12)^2 + (3/12)^2 + (4/12)^2 = 50/144 = 25/72, about 0.347. Without replacement the second draw has one fewer marble of the first color, so the answer is smaller: 19/66, about 0.288.

How do I find the probability of two different colors?

Use the complement. P(different) = 1 - 19/66 = 47/66.

ACT Math: A Bag Has 5 Red, 3 Blue and 4 Green Marbles. Two Are Drawn Without Replacement. P(Same Color)?

The Problem

A bag contains 5 red marbles, 3 blue marbles, and 4 green marbles. Two marbles are drawn at random, one after the other, without replacement. What is the probability that both marbles are the same color?

A) 1/3 B) 19/66 C) 25/72 D) 5/12 E) 19/72

Probability without replacement is a regular on ACT Math. The setup is short; the points are lost on the denominator.

Method 1: Count Pairs (Combinations)

The bag has $5 + 3 + 4 = 12$ marbles. The number of equally likely unordered pairs is

$$ \binom{12}{2} = \frac{12 \cdot 11}{2} = 66 $$

Same-color pairs, color by color:

Red: $\binom{5}{2} = 10$
Blue: $\binom{3}{2} = 3$
Green: $\binom{4}{2} = 6$

Favorable pairs: $10 + 3 + 6 = 19$.

$$ P(\text{same color}) = \frac{19}{66} \approx 0.288 $$

Answer: B

Method 2: Multiply Along the Branches

Add up the three ways to get a matching pair. After one marble of a color is gone, both the count of that color and the total drop by 1:

$$ P = \frac{5}{12}\cdot\frac{4}{11} + \frac{3}{12}\cdot\frac{2}{11} + \frac{4}{12}\cdot\frac{3}{11} = \frac{20 + 6 + 12}{132} = \frac{38}{132} = \frac{19}{66} $$

Both methods agree, which is the best check you can run in under 20 seconds.

Traps on the Answer Choices

C) 25/72: this is the answer if the first marble is replaced: $\left(\frac{5}{12}\right)^2 + \left(\frac{3}{12}\right)^2 + \left(\frac{4}{12}\right)^2 = \frac{50}{144} = \frac{25}{72}$. The problem says without replacement.
A) 1/3: "three colors, so one in three." The colors are not equally likely, so symmetry does not apply.
D) 5/12: this is only the chance that the first marble is red.
E) 19/72: the right numerator over the wrong denominator ($72 = 12 \cdot 6$). The denominator is 66 for unordered pairs, or $12 \cdot 11 = 132$ if you also count order, in which case the numerator doubles to 38.

Two More to Practice (Same Method)

A drawer has 4 black socks and 6 white socks. Two are drawn without replacement. $P(\text{same color}) = \dfrac{\binom{4}{2} + \binom{6}{2}}{\binom{10}{2}} = \dfrac{6 + 15}{45} = \dfrac{7}{15}$.
Same bag as above (5 red, 3 blue, 4 green). $P(\text{both red}) = \dfrac{\binom{5}{2}}{66} = \dfrac{10}{66} = \dfrac{5}{33}$.

The Rule to Remember

If the problem says "without replacement", the second fraction has a smaller denominator and, for the same color, a smaller numerator. If it says "with replacement", both fractions are identical.

For another ACT Math topic that is mostly pattern recognition, see powers of i on the ACT.