The binomial distribution formula helps to check the probability of getting “x” successes in “n” independent trials of a binomial experiment. To recall, the binomial distribution is a type of probability distribution in statistics that has two possible outcomes. In probability theory, the binomial distribution comes with two parameters n and p.
The probability distribution becomes a binomial probability distribution when it meets the following requirements.
- Each trial can have only two outcomes or the outcomes that can be reduced to two outcomes. These outcomes can be either a success or a failure.
- The trails must be a fixed number.
- The outcome of each trial must be independent of each others.
- And the success of probability must remain the same for each trial.
Binomial Distribution Formula in Probability
The formula for the binomial probability distribution is as stated below:
| Binomial Distribution Formula | |
|---|---|
| Binomial Distribution | P(x) = nCx · px (1 − p)n−x |
| Or, | P(r) = [n!/r!(n−r)!]· pr (1 − p)n−r |
Where,
- n = Total number of events
- r (or) x = Total number of successful events.
- p = Probability of success on a single trial.
- nCr = [n!/r!(n−r)]!
- 1 – p = Probability of failure.
Try This: Binomial Distribution Calculator
Examples on Binomial Distribution Formula
Example 1:
A coin is tossed12 times. What is the probability of getting exactly 7 heads?
Solution:
Given that a coin is tossed 12 times. (i.e) n= 12
Thus, a probability pf gettig head in single toss = ½. (i.e) p = ½.
So, 1-p = 1-½ = ½.
We know that the binomial probability distribution is P(r) = nCr · pr (1 − p)n−r.
Now, we have to find the probability of getting exactly 7 heads.(i.e) r = 7.
Substituting the values in the binomial distribution formula, we get
P(7) = 12C7 · (½)7 (½)12−7
P(7) = 792· (½)7 (½)5
P(7) = 792.(½)12
P(7) = 792 (1/4096)
P(7) = 0.193
Therefore, the probability of getting exactly 7 heads is 0.193.
Example 2:
A coin that is fair in nature is tossed n number of times. The probability of the occurrence of a head six times is the same as the probability that a head comes 8 times, then find the value of n.
Solution:
The probability that head occurs 6 times = nC6 (½)6 (½)n-6
Similarly, the probability that head occurs 8 times = nC8 (½)8 (½)n-8
Given that, the probability of the occurrence of a head six times is the same as the probability that a head comes 8 times,
(i.e) nC6 (½)6 (½)n-6 = nC8 (½)8 (½)n-8
⇒nC6(½)n = nC8 (½)n
⇒nC6 = nC8
⇒ 6 = n-8
⇒ n= 14.
Therefore, the value of n is 14.
Example 3:
The probability that a person can achieve a target is 3/4. The count of tries is 5. What is the probability that he will attain the target at least thrice?
Solution:
Given that, p = ¾, q = ¼, n = 5.
Using binomial distribution formula, we get P(X) = nCx · px (1 − p)n−x
Thus, the required probability is: P(X = 3) + P(X=4) + P(X=5)
= 5C3 · (¾)3 (¼ )2 + 5C4 · (¾)4 (¼ )1 +5C5 · (¾)5
= 459/512.
Therefore, the probability that the person will attain the target atleast thrice is 459/512.
Topics Related to Binomial Probability Distribution:
| Probability Distribution Formula | R Squared Formula |
| Skewness Formula | Probability Formulas |
| Quartile Formula | Interquartile Range Formula |
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