BINOMDIST

Calculates probabilities for a binomial distribution.

Syntax:

BINOMDIST(k, n, p, mode)

With n independent trials, each with a probability p of success, BINOMDIST returns the probability that the number of successes will be

exactly k if mode is 0.

up to (and including) k if mode is 1.

In other words, BINOMDIST returns the probability mass function if mode is 0, and the cumulative probability function if mode is 1.

BINOMDIST(k, n, p, 1) is equivalent to B(n, p, 0, k).

Example:

BINOMDIST(3, 12, 0.5, 0)

returns approximately 0.05 (5%), the probability that heads will come up exactly 3 times in 12 flips of a coin.

BINOMDIST(3, 12, 0.5, 1)

returns approximately 0.07 (7%), the probability that heads will come up 0, 1, 2 or 3 times in 12 flips of a coin.

Application:

Manufacturing Quality Control

Scenario: A factory produces light bulbs, and historical data shows that 5% of the light bulbs produced are defective. A quality control inspector randomly selects a sample of 15 light bulbs. We want to find the probability of finding a certain number of defective bulbs in this sample.

Binomial Distribution Parameters:

Number of trials (n): 15 (the number of light bulbs in the sample)
Probability of success (p): 0.05 (the probability of a single light bulb being defective)
Success: A light bulb being defective

Applying the BINOMDIST function:

The BINOMDIST function has four arguments: BINOMDIST(number_s, trials, probability_s, cumulative)

number_s: The number of successes (defective light bulbs) we are interested in.
trials: The total number of trials (15 light bulbs).
probability_s: The probability of a success on each trial (0.05).
cumulative: A logical value. FALSE returns the probability of exactly number_s successes. TRUE returns the probability of at most number_s successes.

Table of Probabilities

We can use the BINOMDIST function to calculate the probability of finding exactly 0, 1, 2, or more defective light bulbs in the sample of 15.

	Number of Defective Bulbs (x)	BINOMDIST Formula (Cumulative = FALSE)	Probability of Exactly * Defective Bulbs
	A	B	C
1	0	BINOMDIST(0, 15, 0.05, FALSE)	0.4633
2	1	BINOMDIST(1, 15, 0.05, FALSE)	0.3658
3	2	BINOMDIST(2, 15, 0.05, FALSE)	0.1348
4	3	BINOMDIST(3, 15, 0.05, FALSE)	0.0307
5	4	BINOMDIST(4, 15, 0.05, FALSE)	0.0049
6	5	BINOMDIST(5, 15, 0.05, FALSE)	0.0006
7	6	BINOMDIST(6, 15, 0.05, FALSE)	0

Interpretation of the results:

The most likely outcome is finding exactly 0 defective light bulbs, with a probability of 46.33%.
There is a 36.58% chance of finding exactly 1 defective bulb.
The probability of finding exactly 2 defective bulbs is 13.48%.
The probability of finding 3 or more defective light bulbs is very low, as shown by the decreasing probabilities. This helps the quality control inspector understand the expected outcomes and identify when a sample might be unusual, potentially indicating a problem with the manufacturing process.

You can also use the cumulative argument set to TRUE to find the probability of at most a certain number of defective bulbs.

	Number of Defective Bulbs (x)	BINOMDIST Formula (Cumulative = FALSE)	Probability of Exactly * Defective Bulbs
	A	B	C
1	0	BINOMDIST(0, 15, 0.05, TRUE)	0.4633
2	1	BINOMDIST(1, 15, 0.05, TRUE)	0.829
3	2	BINOMDIST(2, 15, 0.05, TRUE)	0.9638
4	3	BINOMDIST(3, 15, 0.05, TRUE)	0.9945
5	4	BINOMDIST(4, 15, 0.05, TRUE)	0.9994
6	5	BINOMDIST(5, 15, 0.05, TRUE)	0.9999
7	6	BINOMDIST(6, 15, 0.05, TRUE)	1