## What is Simple Random Sampling?

Simple random sampling is a process in which each article or object in population has an equal chance to get selected and by using this model there are fewer chances of being bias towards some particular objects. There are two ways of sampling in this method a) With replacement and b) Without replacement.

**#1 – Random Sampling with Replacement**

In sampling with replacement, an article once gets selected, then it will be replaced in the population before the next draw. In this way, the same object will have an equal chance to get selected at each draw.

**The formula for “Possible samples with Replacement.” **

There are many different combinations of objects that could be selected while drawing a sample from a population of them.

**No. of possible samples (with replacement) = (Total units)**

^{ (No. of selected units)}**No. of possible samples (with replacement) = N**

^{n }Where,

- N = Number of total Population
- n = Number of units to be selected

For example, let’s assume there is a total of 9 players out of which 3 to be selected to be taken on a playing team, and the selectors decided to use the sample method by replacement.

In that case, there are a number of combination in which players could be selected, i.e.,

N^{n} = 9^{3} = 729

In other words, there are 729 different combinations of three players that could be selected.

**#2 – Random Sampling without Replacement**

In sampling without replacement, an article once gets selected, then it will not be replaced in population. In this way, a particular object will have only a chance to be selected once.

The formula for “Possible samples without Replacement.”

In most commonly used sampling, subjects are typically not included in the sample more than once, i.e., without replacement.

**No. of samples (without replacement) **

**No. of possible samples (without replacement) =**

Where,

- N = Number of people in the population
- n = number of a person to be sampled
- ! = It is the factorial notation

Let’s take the same example, but this time without replacement.

In that case, the number of combination in which players could be selected, i.e.,

- = 9!/3!*(9.3)!
- = 9!/3!*6!
- =9.8.7.6!/3!6!
- =9.8.7/3!
**=84**

In simple words, there are 84 ways to select the combination of 3 players in case of sampling without replacement.

We can see the clear difference in the sample size of the population in case of ‘with replacement’ and ‘without replacement.’

In general, two methods have been used for doing random sampling for a long time. Both of them are as follows:

- Lottery method
- Random number table

**Lottery Method – **This is the oldest method of simple random sampling; in this method, each object in the population has to assign a number & maintain that systematically. Write that number on paper and mix these papers in a box, then numbers are chosen out of the box on a random basis; each number would have the chance to get selected.

**Random Numbers Table – **In this sampling method, it requires to give a number to the population & present that in tabular form; at the time of sampling, each number has the chance to get selected out of the table. Now a day’s software is used for random number table.

### Examples of Simple Random Sampling Formula (with Excel Template)

Let’s further understand the simple random sampling formula by taking examples.

#### Example #1

**If a cinema hall wants to distribute 100 free tickets to its regular customers, the Cinema hall has a list of 1000 number of regular customers in his system. Now the cinema hall can choose 100 customers randomly from its system & can send the tickets to them.**

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**Solution:**

Use the given data for the calculation of simple random sampling.

Calculation of probability(P) can be done as follows:

Probability = No. in the Sample Selected / Total No of Population

- =1000/100

**Probability(P) will be –**

**= 10%**

#### Example #2

**ABC Ltd is a manufacturing company engaged in the manufacturing of bulbs. It manufactures 10 bulbs in a day. It comprises of Quality Inspection team, which is tasked with surprise inspections of bulbs and for measuring the overall feasibility of the company to manufacture Good bulbs. They decided to inspect the bulbs on a random basis, and they decided to take a sample of 3 bulbs, and it was provided that on that particular day, there were 2 defective bulbs and 8 good bulbs. Compare the results in both cases of sampling – with replacement & without replacement.**

**Solution **

Use the given data for the calculation of simple random sampling.

In case of sampling with replacement-

- Number of samples that could be selected = (Total Units)
^{ (}^{No. of selected units of the sample)} - = (10)
^{3} - = 1000

That means there are 1000 possible samples that could be selected.

Let’s denote the population like this – G1, G2, G3, G4, G5, G6, G7, G8, D1, D2.

Then the sample could be (G1, G2, G3), (G1, D1, G7), and so on… Totaling to 1000 samples.

Now let’s say what will be the probability that the sample selected by the invigilator will have at least one of the defective bulbs.

**In case of sampling with replacement**

Probability (at least 1 defective) = Total Probability – Probability (none defective)

Where,

Total Probability means the probability of the total population (universal set), i.e., always 1.

**Calculation of probability of selecting good bulbs**

Probability (none defective) = Probability(Goods) x Probability(Goods) x Probability(Goods)

1^{st} Draw 2^{nd} Draw 3^{rd} Draw

=n(no of good bulbs)/N(Total no of bulbs)*n(no of good bulbs)/N(Total no of bulbs)*n(no of good bulbs)/N(Total no of bulbs)

=8/10*8/10*8/10

**= 0.512**

Now putting these values in the main equation, we will get:

- Probability (at least 1 defective) = Total Probability – Probability (none defective)
- = 1 – 0.512
**= 0.488**

**Explanation** – The probability of selecting Good Bulbs always came 8/10 because, after each draw, the selected bulb was replaced in the Total Group, thus always making the total number of good bulbs in the group 8 and the total size of the group having 10 bulbs in total.

**In case of sampling without replacement**

Probability (at least 1 defective) = Total Probability – Probability (none defective)

**Calculation of probability of selecting good bulbs**

Probability (none defective) = Probability(Goods) x Probability(Goods) x Probability(Goods)

1^{st} Draw 2^{nd} Draw 3^{rd} Draw

=n(no of good bulbs)/N(Total no of bulbs)*n(no of good bulbs)/N(Total no of bulbs)*n(no of good bulbs)/N(Total no of bulbs)

- =8/10*7/9*6/8

**=0.467**

Now putting these values in the main equation, we will get:

Probability (at least 1 defective) = Total Probability – Probability (none defective)

- = 1 – 0.467
**= 0.533**

**Explanation** – The probability of selecting a Good bulb from the group in the 1^{st} draw was 8/10 because, in total, there were 8 good bulbs in the group of a total of 10 bulbs. But after the 1^{st} draw, the selected bulb was not to be selected again, which means it is to be excluded in the next draw. So in the 2^{nd} draw, the Good bulbs were reduced to 7 after excluding the bulb selected in the first draw, and the total bulbs in the group remained 9 making the probability of selecting a Good bulb in the 2^{nd} draw 7/9. The same procedure will be considered for the 3^{rd} draw.

In the given example, you can see that in the case of sampling with replacement, 1^{st}, 2^{nd,} and 3^{rd} draws are independent, i.e., the probability of selecting a good bulb in all the cases would be the same (8/10).

Whereas, in the case of sampling without replacement, each draw is dependent on the previous draw. For example, the probability of selecting a good bulb in the first draw will be 8/10, as there were 8 good bulbs in a total of 10 bulbs. But in the second draw, the number of good bulbs remaining was 7, and the total population size was reduced to 9. Thus the probability became 7/9.

#### Example #3

**Let’s say Mr. A is a Doctor who has 9 patients suffering from a disease for which he has to provide them with regular medications and drug injections, and three of the patient is suffering from Dengue. The record of three weeks is as follows:**

After seeing no results from the medications, the doctor decided to refer them to a specialist Doctor. Due to lack of time, the specialist decided to study 3 patients to examine their conditions and situations.

**Solution: **

To provide an unbiased view of the population, the mean & variance of the sample selected on average be equal to the mean & variance of the whole population, respectively.

Here mean of the population means the average number of the medicines used by the patients in three weeks, which can be calculated by summing up all the no. of injections and dividing it by the total number of patients. (Means forms part of different mathematical concepts as well as in statistics.) * *

**Mean of the Population (X _{p}), **

**Mean of the Population (X _{p}), **

Where,

- Xp = assumed term used for the mean of the population
- Xi = No. of injections for the i
^{th}patient - N = Total number of patients

Putting these values in the equation, we will get

** Calculation of Population Mean**

- Population Mean = (1+2+3+4+5+6+7+8+9)/(9)
**= 10.1 drug injections per patient**

**Explanation** – This means that on average, a patient uses 10.1 drug injections in 3 weeks.

As we can see that in the example, the actual number of injections used by the patients differs from the Mean of the population, we have calculated, and for such a term, *Variance* is used.

Here variance of the population means the average of the square of the difference between the originally used drugs used by the patient and the average drugs used by all the patients (mean of the population).

Population Variance = Sum of Square of difference between actual drugs & average drugs/Total no of patients

= (Actual drug 1st patient -average drug)^2+(Actual drug 2nd patient -average drug)^2 up to 9th patient /total no of patients

= (10-10.1)^2+(8-10.1)^2 ….+(10-10.1)^2 /9

**Calculation of Population Variance**

- = (0.01+4.46+3.57+1.23+0.79+0.79+1.23+0.79+0.01
**Population Variance****= 1.43**

In this case number of the sample that can be selected is = (Total Units)^{ (No. of selected units of the sample)}

^{ }= 9^{3 }= 729

### Relevance and Use

- This process is used to draw conclusions about the population from samples. It is used to determine a population’s characteristics by observing only a portion (sample) of the population.
- Taking a sample requires fewer resources and budget in comparison to observing the whole population.
- A sample will provide needed information quickly while observing the whole population, maybe not feasible, and may take a lot of time.
- A sample may be more accurate than a report on the whole population. A sloppy conducted census can provide less reliable information than a carefully obtained sample.
- In case of an audit, vouching and verification of transactions of a large industry in the given time phrase may not be possible. Hence sampling method is used in such a manner such that an unbiased sample could be selected that represents all of the transactions.

### Recommended Articles

This has been a guide to Simple Random Sampling Formula. Here we discuss the formula for calculation of simple random sampling along with practical examples and a downloadable excel template. You can learn more from the following articles –

- What is Stratified Sampling?
- Calculate Sampling Distribution
- Sampling Error Calculation
- Standard Error
- Effective Annual Rate

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