Binomial Distribution Formula

The binomial distribution is a commonly used discrete distribution in statistics. The normal distribution as opposed to a binomial distribution is a continuous distribution. The binomial distribution represents the probability for 'x' successes of an experiment in 'n' trials, given a success probability 'p' for each trial at the experiment.

Binomial Distribution in Statistics: The binomial distribution forms the base for the famous binomial test of statistical importance. A test that has a single outcome such as success/failure is also called a Bernoulli trial or Bernoulli experiment, and a series of outcomes is called a Bernoulli process. Consider an experiment where each time a question is asked for a yes/no with a series of n experiments. Then in the binomial probability distribution, the boolean-valued outcome the success/yes/true/one is represented with probability p and the failure/no/false/zero with probability q (q = 1 − p). In a single experiment when n = 1, the binomial distribution is called a Bernoulli distribution.

What Is the Binomial Distribution Formula?

The binomial distribution formula is for any random variable X, given by;  P(x:n,p) = ⁿC\(_x\) p^x (1-p)^n-x Or P(x:n,p) = ⁿC_x p^x (q)^n-x

where,

• n = the number of experiments
• x = 0, 1, 2, 3, 4, …
• p = Probability of success in a single experiment
• q = Probability of failure in a single experiment (= 1 – p)

The binomial distribution formula is also written in the form of n-Bernoulli trials, where ⁿC_x = n!/x!(n-x)!. Hence, P(x:n,p) = n!/[x!(n-x)!].p^x.(q)^n-x

Examples on Binomial Distribution Formula

Example 1: If a coin is tossed 5 times, using binomial distribution find the probability of:

(a) Exactly 2 heads

(b) At least 4 heads.

Solution:

(a) The repeated tossing of the coin is an example of a Bernoulli trial. According to the problem:

Number of trials: n=5

Probability of head: p= 1/2 and hence the probability of tail, q =1/2

For exactly two heads:

x=2

P(x=2) = ⁵C2 p² q^5-2 = 5! / 2! 3! × (½)²× (½)³

P(x=2) = 5/16

(b) For at least four heads,

x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5)

Hence,

P(x = 4) = ⁵C4 p⁴ q^5-4 = 5!/4! 1! × (½)⁴× (½)¹ = 5/32

P(x = 5) = ⁵C5 p⁵ q^5-5 = (½)⁵ = 1/32

Answer: Therefore, P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16

Example 2: For the same question given above, find the probability of getting at most 2 heads.

Solution:

Solution: P(at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1)

P(X = 0) = (½)⁵ = 1/32

P(X=1) = 5C1 (½)⁵.= 5/32

Answer: Therefore, P(X ≤ 2) = 1/32 + 5/32 = 3/16

Example 3: 60% of people who purchase sports cars are men. Find the probability that exactly 7 are men if 10 sports car owners are randomly selected.

Solution:
Let’s Identify ‘n’ and ‘X’ from the problem.
The number of sports car owners are randomly selected is n = 10,  and
The number to find the probability is X = 7.

Given: p = 60%, or 0.6. 
Therefore, the probability of failure is q = 1 – 0.6 = 0.4 
Now, using the binomial distribution formula
\( P( x ) = \frac{{n!}}{{( {n - x} )!x!}}.( p )^x .( q)^{n - x}  \\ 
  = \frac{{10!}}{{( {10 - 7} )!7!}}.( {0.6} )^7 .t( {0.4})^{10 - 7}  \\ 
  = 120 \times 0.0279936 \times 0.064 \\ 
  = 0.215 \)
Answer: The probability that exactly 7 are men is 0.215 or 21.5%.