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# Binomial Probability

The arithmetic quantity that tells about the possibility of occurrence of an event is known as probability. If there are totally n feasible ways which are completely and mutually exclusive out of whom in m ways the event A can occur, probability for the event to occur is given by (m / n). In a random sequence of events, n is the trail of events and m are favor to a certain event. By studying in both combined as binomial probability as P(A) = m / n. For an event, if the conclusions can be written as two probabilities, p and q, where p = 1 - q, then the possibility of such an event to occur can be termed an binomial probability.

 Related Calculators Binomial Distribution Probability Calculator Binomial Probability Calculator Binomial Calculator Binomial Confidence Interval Calculator

## Binomial Probability Formula

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The outcomes of the random experiments are called events. The occurrence of an event may be termed as success and non-occurrence can be termed as failure. When there are only two outcomes “success” and “failure” to a random experiment, we can apply Binomial distribution to it.

Binomial Distribution formula is given by

P(x) = nCx px q(n - x),  x = 0, 1, 2, …........, n

Where, p is the probability of success and q is the probability of failure.

1. p lies between 0 and 1. That is, 0 $\leq$ p $\leq$ 1.
2. p + q = 1
3. This formula can be applied to only discrete variables.

### Importance of Binomial Distribution Formula

1. Binomial distribution formula is widely used in quality control.
2. It is used in sampling.
3. It is used in decision-making.
4. It is used to judge the biased of and experiment

### Mean and Standard Deviation of a Binomial Distribution

The mean and standard deviation of Binomial distribution is given by
1. Mean = np
2. Standard Deviation = $\sqrt{npq}$

## Binomial Probability Function

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When only two outcomes are possible as the outcome of an event, the binomial distribution arises. Tossing a coin is an example of implementing binomial probability. Success is represented by number 1 and failure by 0. For the sake of convenience,  the common convention is to use p to represent the probability of the success and q = 1 - p for the probability of failure.The random variable x represents the total success achieved in n number of events for the binomial probability function. It is being represented by P(X = x).

P(X = x) = C(n, x) px q(n - x), where x = 0, 1, 2, …........, n

## Binomial Probability Table

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The binomial table given here shows the most common probabilities listed for n. Here. the probability p of success is used for each x, and the probability is given for obtaining at most x successes in n independent trials

P(X $\leq$ x) = $\sum_{r = 0}^x C(n, r) p^r(1 - p)^{n-r}$ ## Calculate Binomial Probability

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In order to calculate the Binomial probability, it is important to understand and be able to apply the formula of Binomial probabilities. Let us study with the help of example, how to calculate the binomial distribution.

### Solved Example

Question: Suppose a die is tossed 7 times. What is the probability of getting exactly 3 fours?
Solution:

General Formula for finding solution

P(X = r) = (ncr) pr(1 - p)(n - r)

Step 1:

Number of trials, n = 7

Number of success, r =3

Probability of success in any single trial p is given as $\frac{1}{6}$ or 0.167

Step 2:

To calculate nCr formula is used.

nCr = 35

Step 3:

Find pr.

pr = (0.167)3

= 0.004657463

Step 4:

To Find (1- p)n-r Calculate 1 - p and n - r.

1 - p = 1 - 0.167 = 0.833

n - r = 7 - 3 = 4

Step 5:

Find (1 - p)n-r.

= (0.833)4 =0.481481944

Step 6:

Solve P(X = r) = nCr p r (1-p)n-r

= 10 × 0.004657463 × 0.481481944

= 0.0224248434

The probability of getting exactly 3 fours is 0.0224248434

## Binomial Probability Examples

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Given below are some of the examples on Binomial Probability.

### Solved Examples

Question 1: Assume that there are taking about 10 question multiple choice test. If each question has four choices and you have to guess on each question, what is the probability of getting exactly 7 questions correct using binomial Probability formula?
Solution:

Step 1:
Total number of questions, n = 10
Correct answers, k = 7
Therefore, wrong answers are, n – k = 3

Step 2:
p = 0.25 = probability of guess the correct answer on a question
q = 0.75 = probability of guess the wrong answer on a question

=> P(7 Correct guesses out of 10 questions) = 0.0031 approximate value.

Therefore if someone guesses 10 answers on a multiple choice test with 4 options, they have about a 5.8% chance of getting 5 and only 5 correct answers. If 5 or more correct answers are needed to pass then probability of passing can be calculated by adding the probability of getting 5 (and only 5) answers correct, 6 (and only 6) answers correct, and so on up to 10 answers correct. Total probability of 5 or more correct answer is approximate percentage is 7.8

Question 2: Eight unbiased coins were tossed simultaneously. Find the probability of getting
(i) exactly 4 heads (ii) no heads at all (iii) 6 or more heads (iv) at most two heads (v) number of heads ranging from 3 to 5.
Solution:

Step 1:
Total number of coins are, n = 8.

The event of getting head is a success and event of getting tail is a failure. The probability of getting a head and getting a tail is equally likely.

=> Probability of getting a head = Probability of getting a tail = $\frac{1}{2}$.

That is p = q = $\frac{1}{2}$

Step 2:
Binomial Distribution formula is given by

P(x) = nCx px qn-x, where x = 0, 1, 2, …n

Let x be the number of heads

Case 1:
(i) Exactly 4 heads. That is x = 4

Substituting x = 4 in the formula

=> P(Exactly 4 heads) = p(x = 4)

= $8c^{4}\left ( \frac{1}{2} \right )^4\left ( \frac{1}{2} \right )^{8-4}$

= $8c_{4}\left ( \frac{1}{2} \right )^4\left ( \frac{1}{2} \right )^4$

= $\frac{70}{256}$

Case 2:
(ii) No heads at all . That is x = 0

Substituting x = 0 in the formula

=> P(No heads at all) = p(x = 0) = $8c_{0}\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^{8 - 0}$

= $8c_{0}\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^8$

= $\frac{1}{256}$

Case 3:
(iii) 6 or more heads. That is x = 6, 7 or 8

=> P(6 or more heads) = p(x = 6) + p(x = 7) + p(x = 8)

= $8c_6\left ( \frac{1}{2} \right )^6\left ( \frac{1}{2} \right )^{8-6}+8c_7\left ( \frac{1}{2} \right )^7\left ( \frac{1}{2} \right )^{8-7}8c_8\left ( \frac{1}{2} \right )^8\left ( \frac{1}{2} \right )^{8-8}$

= $8c_6\left ( \frac{1}{2} \right )^6\left ( \frac{1}{2} \right )^2+8c_7\left ( \frac{1}{2} \right )^7\left ( \frac{1}{2} \right )^1+8c_8\left ( \frac{1}{2} \right )^8\left ( \frac{1}{2} \right )^0$

= $\frac{28}{256}+\frac{8}{256}+\frac{1}{256}$

= $\frac{37}{256}$

Case 4:

(iv) At most two heads. That is x = 0, 1 or 2

P(at most two heads) =  p(x = 0) + p(x = 1) + p(x = 2)

= $8c_0\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^{8-0}+8c_1\left ( \frac{1}{2} \right )^1\left ( \frac{1}{2} \right )^{8-1}+8c_2\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^{8-2}$

= $8c_0\left ( \frac{1}{2} \right )^0\left ( \frac{1}{2} \right )^8+8c_1\left ( \frac{1}{2} \right )^1\left ( \frac{1}{1} \right )^7+8c_2\left ( \frac{1}{2} \right )^2\left ( \frac{1}{2} \right )^6$

= $\frac{1}{256}+\frac{8}{256}+\frac{28}{256}$

= $\frac{37}{256}$

Case 5:
(v) Number of heads ranging from 3 to 5 That is x = 3, 4 or 5

=> P(Number of heads ranging from 3 to 5) = P(x = 3) + P(x = 4) + P(x = 5)

= $8c_3\left ( \frac{1}{2} \right )^8\left ( \frac{1}{2} \right )^{8-3}+8c_4\left ( \frac{1}{2} \right )^4\left ( \frac{1}{1} \right )^{8-4}+8c_5\left ( \frac{1}{2} \right )^5\left ( \frac{1}{2} \right )^{8-5}$

= $8c_3\left ( \frac{1}{2} \right )^3\left ( \frac{1}{2} \right )^5+8c_4\left ( \frac{1}{2} \right )^4\left ( \frac{1}{1} \right )^4+8c_5\left ( \frac{1}{2} \right )^5\left ( \frac{1}{2} \right )^3$

= $\frac{56}{256}+\frac{70}{256}+\frac{56}{256}$

= $\frac{182}{256}$
.

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