Cumulative Binomial Probability Table

The cumulative binomial table gives P(X ≤ k): the probability of k or fewer successes in n independent trials, each with success probability p. Find the block for your n, the row for k and the column for p.

Turning the table into other probabilities

• Exactly k: P(X = k) = P(X ≤ k) − P(X ≤ k − 1). For 10 tosses, P(X = 6) = 0.8281 − 0.6230 = 0.2051.
• At least k: P(X ≥ k) = 1 − P(X ≤ k − 1). A quality inspector samples 15 items from a line with a 20% defect rate; the chance of finding 6 or more defects is 1 − 0.9389 = 0.0611.
• More than k: P(X > k) = 1 − P(X ≤ k).
• Between a and b (inclusive): P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1).

To check a single probability or use values of n and p not listed here, use the binomial calculator.

Jump to n

n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

n = 7

n = 8

n = 9

n = 10

n = 11

n = 12

n = 13

n = 14

n = 15

n = 16

n = 17

n = 18

n = 19

n = 20

Computed with SciPy (binom.cdf) and rounded to four decimals. A value shown as 1.0000 is at least 0.99995 but, unless k = n, not exactly 1.

When the binomial model applies

The binomial distribution needs four conditions: a fixed number of trials n; each trial has two outcomes (success or failure); the trials are independent; and the success probability p is the same on every trial. Drawing cards without replacement breaks independence and calls for the hypergeometric distribution instead; counting events in a time period with no fixed n calls for the Poisson distribution.

Using the table for a binomial test

Suppose a supplier claims at most 10% of parts are faulty, and a sample of 20 contains 4 faulty parts. Is that evidence against the claim? From the block for n = 20, column p = 0.10, P(X ≤ 3) = 0.8670, so P(X ≥ 4) = 1 − 0.8670 = 0.1330. Seeing 4 or more faulty parts would happen about 13% of the time even if the claim were true, so it is not strong evidence against it at the 5% level.

Frequently asked questions

How do you read a cumulative binomial table?

Find the block for the number of trials n, the row for k and the column for the success probability p. The value is the probability of k or fewer successes.

How do I get P(X = k) from a cumulative table?

Subtract consecutive entries: P(X = k) equals P(X <= k) minus P(X <= k - 1).

How do I find the probability of at least k successes?

Use 1 minus P(X <= k - 1). For example, at least 6 successes is 1 minus the table value at k = 5.

What if my p is not in the table?

Use the binomial calculator, which accepts any n and p. For p above 0.5 you can also count failures instead of successes, since failures follow a binomial with probability 1 - p.

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