How to Find Probabilities in a Box of Red and Black Balls

A question about a box of red and black balls usually asks for the chance of drawing a particular color mix. The answer depends on how many balls of each color are in the box, how many you draw, and whether you put each ball back before the next draw.

The standard case is a draw without replacement: a ball stays out of the box after you draw it. This article gives the formula for that case, works through an example, and explains how the answer changes when you replace the balls.

The formula for drawing without replacement

Suppose the box contains $R$ red balls and $B$ black balls, and you draw $n$ balls. If you want exactly $k$ red balls, then you must also draw $n-k$ black balls. The probability is

Here, $\binom{a}{b}$ means the number of ways to choose $b$ objects from $a$ when order does not matter. The numerator counts the favorable selections: choose $k$ of the $R$ red balls and $n-k$ of the $B$ black balls. The denominator counts every possible selection of $n$ balls from the whole box.

The formula applies when the requested counts are possible: $0\leq k\leq R$ and $0\leq n-k\leq B$. If either condition fails, the probability is zero.

Worked example: exactly two red balls

A box contains 5 red balls and 7 black balls. You draw 4 balls without replacement. What is the probability that exactly 2 are red?

There are $\binom{12}{4}$ ways to select any 4 balls. A favorable selection has 2 of the 5 red balls and 2 of the 7 black balls:

So the probability is $14/33$, or about 42.4%. This method counts sets of balls, not sequences. For instance, it does not count “red, then black, then red, then black” separately from other orders; all orders producing the same selected group are already included in the combinations.

Probabilities such as “at least one red”

For a range of possible counts, add the probabilities for the allowed values of $k$. For example, “at least one red” means 1, 2, 3, or more red balls, up to the number drawn. Often it is simpler to use the complement: find the probability of drawing no red balls and subtract it from 1.

In the same box, the chance of drawing 4 black balls is $\binom{7}{4}/\binom{12}{4}$. Therefore,

This complement approach is especially useful when the event includes many possible color mixes but its opposite is easy to count.

What changes if you replace each ball?

If you put each ball back before the next draw, the composition stays the same on every draw. The probability of red on any draw is $p=R/(R+B)$, and the draws are independent. For exactly $k$ red balls in $n$ draws, use the binomial formula:

For example, with 5 red and 7 black balls, the chance of exactly 2 red in 4 draws with replacement is

This differs from the no-replacement result, $14/33\approx0.4242$, because drawing a ball without replacement changes the color proportions remaining in the box.

Common mistakes

• Ignoring replacement: Decide whether a drawn ball goes back. It determines which model applies.
• Using permutations when order is irrelevant: If the question asks only for the final color counts, combinations are the natural count. If it asks for a specific sequence, calculate the sequence probability instead.
• Forgetting the complementary color: Exactly $k$ red in a draw of $n$ means exactly $n-k$ black.
• Treating balls of the same color as one outcome: The combinations count the distinct physical balls, even though balls are described only by color. If instead every red/black sequence is treated as equally likely, account for the changing probabilities under no replacement; sequences generally are not equally likely.

For draws without replacement, the count of red balls in the sample follows a hypergeometric distribution. The combinations formula is that distribution’s probability mass function, expressed directly in terms of the numbers of balls in the box.