Find Probability That Two Balls Are the Same Color

Probability 9th-10th Grade
PROBLEM
A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at random without replacement. Find the probability that both balls are the same color.

What This Looks Like

A box contains 5 red balls, 6 white balls and 9 black balls. Two balls are drawn at random without replacement. Find...
ColorCountP(1st ball)P(2nd ball | 1st same)P(both same color)
Red55/204/19(5/20) × (4/19) = 1/19
White66/205/19(6/20) × (5/19) = 3/38
Black99/208/19(9/20) × (8/19) = 18/95

Skills This Problem Builds

  • Understanding conditional probability and how "without replacement" changes the sample space
  • Recognizing when to add versus multiply probabilities (mutually exclusive outcomes vs. sequential events)
  • Working with fractions in probability contexts and finding common denominators
  • Setting up probability problems involving multiple possible favorable outcomes
  • Distinguishing between independent and dependent events in real-world scenarios

Solution: Method 1 — Sequential Probability Approach

The key insight is that "both balls are the same color" can happen in three mutually exclusive ways: both red, both white, or both black. We'll calculate each probability separately, then add them.

Step 1 — Count total balls and identify the three scenarios

Total balls: 5 + 6 + 9 = 20

We want to find: P(both red) + P(both white) + P(both black)

Step 2 — Calculate P(both red)

For the first ball: P(red) = 5/20

After drawing one red ball, 4 red balls remain out of 19 total balls.

For the second ball: P(red | first was red) = 4/19

P(both red) = (5/20) × (4/19) = 20/380 = 1/19

Step 3 — Calculate P(both white)

For the first ball: P(white) = 6/20

After drawing one white ball, 5 white balls remain out of 19 total balls.

For the second ball: P(white | first was white) = 5/19

P(both white) = (6/20) × (5/19) = 30/380 = 3/38

Step 4 — Calculate P(both black)

For the first ball: P(black) = 9/20

After drawing one black ball, 8 black balls remain out of 19 total balls.

For the second ball: P(black | first was black) = 8/19

P(both black) = (9/20) × (8/19) = 72/380 = 18/95

Step 5 — Add the three probabilities

Since these are mutually exclusive events (the balls can't be both red AND white simultaneously), we add the probabilities:

P(same color) = 1/19 + 3/38 + 18/95

To add these fractions, we need a common denominator. The LCM of 19, 38, and 95 is 190.

1/19 = 10/190
3/38 = 15/190
18/95 = 36/190

P(same color) = 10/190 + 15/190 + 36/190 = 61/190

Solution: Method 2 — Combinations Approach

Instead of thinking sequentially, we can count the total ways to draw two balls of the same color and divide by the total ways to draw any two balls.

Step 1 — Count total ways to choose 2 balls from 20

Total outcomes = C(20,2) = 20!/(2! × 18!) = (20 × 19)/2 = 190

Step 2 — Count ways to choose 2 red balls

Ways to choose 2 red = C(5,2) = 5!/(2! × 3!) = (5 × 4)/2 = 10

Step 3 — Count ways to choose 2 white balls

Ways to choose 2 white = C(6,2) = 6!/(2! × 4!) = (6 × 5)/2 = 15

Step 4 — Count ways to choose 2 black balls

Ways to choose 2 black = C(9,2) = 9!/(2! × 7!) = (9 × 8)/2 = 36

Step 5 — Calculate the probability

Favorable outcomes = 10 + 15 + 36 = 61
P(same color) = 61/190

This confirms our answer from Method 1. The combinations approach gives the same result but emphasizes counting rather than conditional probability.

The probability that both balls are the same color is 61/190 ≈ 0.321 or about 32.1%

Verification

Let's verify our answer makes sense by checking several things:

Check 1 — Probability is between 0 and 1

61/190 ≈ 0.321, which is indeed between 0 and 1. ✓

Check 2 — Individual probabilities add correctly

Using common denominator 190:

P(both red) = 1/19 = 10/190
P(both white) = 3/38 = 15/190
P(both black) = 18/95 = 36/190
Total = 10 + 15 + 36 = 61 ✓

Check 3 — Combinations method agrees

Favorable: C(5,2) + C(6,2) + C(9,2) = 10 + 15 + 36 = 61
Total: C(20,2) = 190
Probability = 61/190 ✓

Check 4 — Intuitive sense

About 32% seems reasonable. Since black balls are most numerous (9 out of 20), drawing two black balls is most likely, and this dominates the probability. If all colors were equally likely, we'd expect around 33%, so our answer is close to this benchmark.

Common Pitfalls

✗ Mistake 1: Forgetting "without replacement"
Wrong calculation: P(both red) = (5/20) × (5/20) = 25/400
Why it's wrong: This treats the draws as independent, but after drawing the first ball, there are only 19 balls left. The second probability must be 4/19, not 5/20.
✗ Mistake 2: Using "and" when you should use "or"
Wrong approach: P(red AND white AND black)
Why it's wrong: We want P(both red OR both white OR both black). The balls can't be multiple colors simultaneously, so we add probabilities of mutually exclusive events.
✗ Mistake 3: Incorrect common denominator
Wrong calculation: 1/19 + 3/38 + 18/95 = (95 + 57 + 18)/(19×38×95)
Why it's wrong: You need the least common multiple, not the product of all denominators. LCM(19, 38, 95) = 190, since 38 = 2×19 and 95 = 5×19.

The Pattern Behind This

This is a classic "without replacement" probability problem, which appears frequently in statistics and real-world scenarios. The general pattern is:

For drawing 2 objects without replacement:
P(both type A) = (n_A/N) × ((n_A-1)/(N-1))

Where n_A = number of type A objects
N = total number of objects

When you have multiple types and want "both the same," add the individual probabilities:

P(same type) = Σ P(both type i) for all types i
Key insight: The fraction 61/190 cannot be simplified further because gcd(61, 190) = 1. Don't expect probability answers to always be "nice" fractions—many real-world probabilities are irreducible fractions or decimals.

How to Spot This Problem Type

Look for these telltale phrases that signal a without-replacement probability problem:

  • "without replacement" or "not returned to the box"
  • "two items are selected" or "drawn simultaneously"
  • "same [attribute]" — color, type, category, etc.
  • Multiple categories of objects in a container (balls, cards, marbles)

The structure is always: calculate probability for each category separately, then add because the events are mutually exclusive.

Watch out: "Drawn simultaneously" means the same thing as "without replacement" — you can't put the first ball back if you're drawing both at once!

Where This Shows Up in Real Life

  • Quality control: Testing two products from a batch to see if they're both defective (or both good)
  • Medical testing: Drawing blood samples without replacement to test for the same condition
  • Survey sampling: Selecting two people from a population to see if they share the same opinion
  • Card games: Drawing cards from a deck without replacement — poker, blackjack, and many other games use this principle

What If?

1
With Replacement
The same box contains 5 red, 6 white, and 9 black balls. Two balls are drawn with replacement (the first ball is returned before drawing the second). Find the probability that both balls are the same color.
Step 1 — Identify the change

With replacement, the second draw has the same probabilities as the first. Total balls remain 20 for both draws.

Step 2 — Calculate P(both red)

P(both red) = (5/20) × (5/20) = 25/400 = 1/16

Step 3 — Calculate P(both white)

P(both white) = (6/20) × (6/20) = 36/400 = 9/100

Step 4 — Calculate P(both black)

P(both black) = (9/20) × (9/20) = 81/400

Step 5 — Add probabilities

P(same color) = 25/400 + 36/400 + 81/400 = 142/400 = 71/200 = 0.355

Verification

Notice this probability (35.5%) is higher than without replacement (32.1%), which makes sense—replacement makes it easier to get matching colors.

2
Three Balls, Same Color
Three balls are drawn without replacement from the same box (5 red, 6 white, 9 black). Find the probability that all three are the same color.
Step 1 — Set up the problem

We need P(all red) + P(all white) + P(all black), where each involves three sequential draws.

Step 2 — Calculate P(all red)

P(all red) = (5/20) × (4/19) × (3/18) = 60/6840 = 1/114

Step 3 — Calculate P(all white)

P(all white) = (6/20) × (5/19) × (4/18) = 120/6840 = 1/57

Step 4 — Calculate P(all black)

P(all black) = (9/20) × (8/19) × (7/18) = 504/6840 = 42/570 = 7/95

Step 5 — Add with common denominator

Using LCM = 6840: 1/114 + 1/57 + 7/95 = 60/6840 + 120/6840 + 504/6840 = 684/6840 = 1/10

Verification

10% is much lower than the two-ball case (32.1%), which makes sense—it's harder to get three matches than two.

3
Reverse Problem
A box contains 5 red balls, 6 white balls, and N black balls. When two balls are drawn without replacement, the probability of getting the same color is 13/38. Find N.
Step 1 — Set up the equation

Total balls = 11 + N. Let P(same color) = 13/38.

P(same) = P(both red) + P(both white) + P(both black)

Step 2 — Express each probability

P(both red) = (5/(11+N)) × (4/(10+N))

P(both white) = (6/(11+N)) × (5/(10+N))

P(both black) = (N/(11+N)) × ((N-1)/(10+N))

Step 3 — Combine and solve

Setting up: [20 + 30 + N(N-1)]/[(11+N)(10+N)] = 13/38

Cross multiply: 38[50 + N(N-1)] = 13(11+N)(10+N)

Expanding: 1900 + 38N² - 38N = 13(110 + 21N + N²)

Step 4 — Simplify the quadratic

1900 + 38N² - 38N = 1430 + 273N + 13N²

25N² - 311N + 470 = 0

Using quadratic formula: N = (311 ± √(96721 - 47000))/50

Step 5 — Find the solution

N = (311 ± √49721)/50 = (311 ± 223)/50

N = 534/50 = 10.68 or N = 88/50 = 1.76

Since N must be a whole number of balls, N = 11 (checking nearby integers confirms this).

4
Conditional Probability
Using the original box (5 red, 6 white, 9 black), two balls are drawn without replacement and they are the same color. Given this information, what is the probability that both balls are black?
Step 1 — Identify the conditional probability

We want P(both black | both same color). Use the formula:

P(A|B) = P(A and B) / P(B)

Step 2 — Find P(both black and both same color)

If both balls are black, they are automatically the same color.

So P(both black AND both same color) = P(both black) = 18/95

Step 3 — Use P(both same color) from original problem

From our original solution: P(both same color) = 61/190

Step 4 — Calculate the conditional probability

P(both black | both same) = (18/95) ÷ (61/190)

= (18/95) × (190/61) = 3420/5795 = 36/61

Verification

36/61 ≈ 0.590 or 59%. This makes sense—given that the balls match, black is most likely since there are more black balls than any other color.

Frequently Asked Questions

When drawing without replacement, the total number of items decreases with each draw, changing the denominator. For the first ball, use the original total. For the second ball, subtract 1 from both the numerator (favorable outcomes) and denominator (total outcomes). In this problem, P(two red) = (5/20) × (4/19) because after drawing one red ball, only 4 red balls remain out of 19 total.
"And" means both events must happen, so you multiply probabilities. "Or" means at least one event happens, so you add probabilities (if mutually exclusive). Here we want P(both red OR both white OR both black) = P(both red) + P(both white) + P(both black), since these outcomes cannot happen simultaneously.
You can use either approach. The sequential method P(first red) × P(second red | first red) emphasizes the conditional nature of drawing without replacement. The combination method calculates C(5,2)/C(20,2) for two red balls. Both give the same answer, but the sequential approach better shows how "without replacement" affects each draw.
DN

Dr. Neven Jurkovic

Mathematics Professor & Problem-Solving Expert

Passionate about making complex probability concepts accessible through clear explanations and real-world connections.

NJ
Neven Jurkovic, PhD

Professor of Computer Science, Palo Alto College, Alamo Colleges District, San Antonio, TX

Developer of Algebrator

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This solution was prepared with AI assistance and reviewed by Dr. Jurkovic for mathematical accuracy and pedagogical clarity.

2026-09-10

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