Binomial Distribution Calculator

The binomial distribution calculator, also known as binomial calculator, or binomial probability calculator is an online tool developed to help you compute probabilities for binomial experiments. Unlike many tools that provide only numerical results, our free binomial probability calculator provide you with a step-by-step solution for P(X=k). This feature makes it particularly valuable for students seeking to understand probability concepts, as well as for instructors, researchers, and professionals who require accurate results. Additionally, the calculator provides the mean and standard deviation, as well as the corresponding probability mass function (PMF). Thus, with our binomial probability calculator, you will gain both efficiency and deeper insight into the binomial distribution.

Solved Example: How to Calculate Binomial Probabilities

A light bulb manufacturer knows that about 3% of bulbs are defective. A quality inspector randomly tests 20 bulbs from a shipment. Let X be the number of defective bulbs found.

Answer the following questions:

i) What is the probability that at most 2 bulbs are defective?
ii) What is the probability that exactly 2 bulbs are defective?
iii) What is the probability that at least 3 bulbs are defective?
iv) What is the probability that less than 2 bulbs are defective?
v) What is the probability that greater than 1 bulbs is defective?

From the problem, we know that

• n = 20 is the number of trials
• p = 0.03 is the probability of success, i.e. defective bulb
• q = 1-p = 0.97 is the probability of failure, i.e. non-defective bulb

The random variable (X) represents the number of defective bulbs out of 20.

(i) What is the probability that at most 2 bulbs are defective?

This question requires us to compute P(X≤2)

Thus, n = 20, p = 0.03, and k = 2

To solve the problem manually, we follow this procedure:

(i) At most 2

$P(X\le2) = P(X=0)+P(X=1)+P(X=2)$

$P(X=0) = \binom{20}{0}(0.03)^0(0.97)^{20}$

$P(X=1) = \binom{20}{1}(0.03)^1(0.97)^{19}$

$P(X=2) = \binom{20}{2}(0.03)^2(0.97)^{18} $

Thus:

$P(X \le 2) = 0.543794 + 0.336368 + 0.098830$

$= \mathbf{0.978992} $

Now using our binomial distribution calculator, simply enter the number of trials (n = 20), probability of success (p = 0.03), and the number of successes (k = 2). Then hit the “calculate” button:

You’ll get the following outputs:

As you can see, the required probability is 0.9790, which coincide with the manual calculation.

ii) What is the probability that exactly 2 bulbs are defective?

This question requires us to compute p (X=2)

Thus, n = 20, p = 0.03, and k = 2

To solve the question manually, we have:

$P(X=2) = \binom{20}{2}(0.03)^2(0.97)^{18}$

$= 190 \times (0.03)^2 \times (0.97)^{18}$

$\approx \mathbf{0.098830}$

Now, using our binomial probability calculator, you simply need to insert n = 20, p = 0.03, and k =2 and look for the P(X=2) from the results, as shown below:

Now, you can see that both the manual computation and our free calculator yield the same values. In other words, P(X = 2) = 0.0988 in both cases.

iii) What is the probability that at least 3 bulbs are defective?

Here, we need to find P(X≥3)

Thus, n =20, p = 0.03, and k = 3

Substituting the values in the binomial distribution probability formula, we have:

$P(X \ge 3) = 1 – P(X \le 2)$

$= 1 – (P(X=0)+P(X=1)+P(X=2))$

But from (i), we know that:

P(≤2) = 0.978992

Thus, $P(X \ge 3) =1- 0.978992$

$\mathbf{0.021008}$

Now, using our online binomial probability calculator, we simply input n = 20, p = 0.03, and k =3. Next, hit the “Calculate” button to see the results. You can then read the required probability from the results (i.e. P(X≥3)) as shown below:

As you’ll note, the manual calculation and the calculator still yield the same results (0.0210). This implies that you can always use our binomial calculator to verify manual calculation results.

iv) What is the probability that less than 2 bulbs are defective?

This question requires you to calculate P(X < 2). This is equivalent to finding P(X≤1).

From the question, we know that: n = 20, p = 0.03, and k =2

And P(X≤1) = P(X=0) + P(X=1)

From (i), P(X=0) =0.543794 and P(X=1) = 0.336368

Thus, the required P(X<2) is given by:

$P(X<2) = 0.543794 + 0.336368$

$= \mathbf{0.880162}$

Using our binomial distribution calculator, we only need to specify the given values. That is, n = 20, p =0.03, and k =2. Next, read the values corresponding to P(X<2), as shown below:

The calculator still show results that are consistent with manual calculation. In other words, both manual calculation and our binomial probability calculator reveal that P(X<2) = 0.8802.

v) What is the probability that greater than 1 bulbs is defective?

Here, we need to find P(X>1). But we know that n = 20, k =1, and p = 0.03

P(X>1) can be redefined as follows:

P(x>1) = 1−P(X≤1))

Where, P(X≤1) = P(X=0) + P(X=1)

But from (iv), P(X≤1) = 0.880162

Thus:

$P(X > 1) = 1 – 0.880162$

$= \mathbf{0.119838}$

Using Our Free Binomial Distribution Probability Calculator, simply specify, n = 20, p = 0.03, and k = 1. Once you hit the “Calculate” button, you’ll only need to read P(X>1), as shown below:

Just as the manual computation using the binomial formula, our free binomial calculator yielded the same findings. That is, P(X>1) = 0.1198.

Frequently Asked Questions