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Random Sampling is a statistical technique for selecting a subset of the population from which to make statistical inferences. As a result, the sampling method enables researchers to gather information about a whole population based on data from a subset.
- Random sampling is also known as Probability sampling.
- It starts with a full list of eligible candidates who all have an equal chance of being selected.
- The selection must be random and no different from the individuals who were not sampled.
- Random sampling reduces the number of chances of mistakes.
- The method requires little advance knowledge regarding the population.
- It help in predicting the outcome of any particular event.
- Purchasing fruits from a vendor is a real-life example of the sampling.
Read More: Statistics
Key Terms: Random Sampling, Probability, Sampling, Simple Random Sampling, Systematic Random Sampling, Stratified Random Sampling, Clustered Sampling
Random Sampling
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Random sampling is a technique for selecting a random sample of observations from a population to make assumptions about the population. Non-probability sampling, also known as non-random sampling, is the inverse of this sampling method.
- Simple random sampling, stratified sampling, cluster sampling, and multistage sampling are the most common types of sampling.
- Non-arbitrary samples are commonly referred to as convenience samples in sampling methods.
- It derives assumptions about the entire population.
- Random sampling takes a random portion of the entire population.
- The benefit of using probability sampling is that it ensures that the sample is representative of the general population.
- It is a time-consuming and costly procedure.
Read More: Difference between SD and Variance
Solved Example of Random SamplingExample: Assume a company has 1000 employees, of which 100 are required to complete onsite work. All of their names are now in the basket, and 100 will be chosen at random. In this case, each employee has an equal chance of being selected. Solution: Once the sample size and population are known, one can easily pick the probability from this dataset. Here's the equation: The possibility of a one-time selection is as follows: P = n/N = 100/1000 = 10 % Also, more than once: P = 1-(1-(1/N))n P = 1 – (999/1000)n P = 0.952 P = 9.5 % |
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Types of Random Sampling
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The random sampling method employs some form of random selection. In this method, all suitable individuals can select a sample from the entire sample space. Here are some of the most important types of random sampling techniques are as follows:
Simple Random Sampling
Simple random sampling is the easiest way to obtain random samples. It entails selecting the desired sample size and selecting observations from people so that everyone has an equal chance of being chosen until the final sample size is determined.
Solved Example of Simple Random SamplingExample: A random selection of 20 students from a total of 50 in a single class, for example, provides a 1/50 chance of being chosen. |
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Systematic Random Sampling
In systematic random sampling, the items are chosen from the destination population. It first determines the random selecting point and then uses the other methods after a fixed sample period. It equals the ratio of the total population size to the required population size.
Stratified Random Sampling
Stratified random sampling divides the entire database into significant subgroups or strata. Furthermore, the elements are chosen at random from each stratum. The required sample size will now have a design corresponding to the population size or representing its sub-categories.
- The primary advantage is that it allows for a more focused approach to sample selection.
Solved Example of Simple Random SamplingExample: If a sample size of 200 is required and there are four groups to choose from, choosing 50 samples from every group will be enough. |
Clustered Sampling
Cluster sampling is similar to stratified sampling in that the population is divided into many subgroups (for example, hundreds of thousands of strata or subgroups). A few of these subgroups are selected at random, and simple random samples are collected within these subgroups.
- These subgroups are referred to as clusters.
- It is primarily used to reduce the cost of data compilation.
Read More: Population and Sampling
Random Sampling Formula
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The random sampling formula is given as
P = 1 – (N-1/N)(N-2/N-1)....(N-n/N-(n-1)).
- Here, P is the probability, n is the sample size, and N is the population.
- When 1-(N-n/n) is canceled, P = n/N is obtained.
- Furthermore, the probability of a sample being chosen more than once is required: P = 1-(1-(1/N)) n.
Read More: Multiplication Theorem on Probability
Advantages of Simple Random Sampling
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The advantages of random sampling in research are as follows:
- Random sampling helps reduce the bias in the sample.
- It is considered a fair method of sampling.
- This method does not necessitate any technical knowledge.
- It is a fundamental method of data collection.
- The data generated using this method is accurate.
- Researchers can create any sample size.
- The size of the population is large in the simple random sampling method.
- It is simple to select a smaller sample size from a larger population.
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Things to Remember
- Random Sampling is a selection of random samples of participants to derive conclusions.
- It helps in selecting a subset of the population from which to make statistical inferences.
- The method enables researchers to gather information about a whole population based on data from a subset.
- Random sampling is required to draw an unbiased conclusion from a large data pool.
- It reduces the possibility of error and expedites the process.
- There are four types of Random Sampling: Simple Random Sampling, Systematic Random Sampling, Stratified Random Sampling and Cluster Sampling.
- It is simple to choose a smaller sample size from a larger population.
Also read: Differentiation and Integration Formula
Sample Questions
Ques. How are Random samples chosen? Describe the sample space for the experiment: A coin is tossed and a dice is thrown. (4 marks)
Ans: Random samples are drawn from an exhaustive list of a large population and then chosen at random. Because of this method, every entity in a large data pool has an equal chance of being chosen. Typically, researchers use one of 2 methods to manage this process: one is a manual lottery, and the other is to draw them at random from a sample group.
- A coin is tossed, and a die is thrown. A coin has 2 faces: the head (H) and the tail (T).
- A die has 6 faces, each numbered from 1 to 6, and each face has one number.
- Thus, when a coin and a die are tossed, the sample space is given by: S = (H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6)
Ques. A box contains 1 red ball and 3 identical white balls. 2 balls are drawn at random, one after the other, with no replacement. Write the sample space for this experiment. (2 marks)
Ans: The box contains 1 Red ball and 3 identical white balls. Let us use the letter R to represent the red ball and the letter W to represent the white ball. The sample space is given by S= when two balls are drawn at random in succession without replacement (RW, WR, WW).
Ques. An experiment entails tossing a coin and then tossing it again if it lands on its head. If the first toss results in a tail, a die is rolled once more. Find the sample area. (2 marks)
Ans: A coin has 2 faces: the head (H) and the tail (T) (T). A die has 6 faces, each of which is numbered from 1 to 6 and has one number on it. As a result, the sample space in the given experiment is given by S= (HH, HT, T1, T2. T3, T4, T5, T6).
Ques. A die is thrown repeatedly until a 6 comes up. What is the sample space for the experiment? (2 marks)
Ans: 6 may appear on the first throw, the second throw, the third throw, and so on until 6 is obtained in this experiment. Therefore, the sample space of experiment is given by S = (6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), (1, 5, 6), (2, 1, 6), (2, 2, 6), .... (2, 5, 6), (5, 1, 6), (5, 2, 6)....)
Ques. Determine the probability that a hand of 7 cards drawn from a well-shuffled deck of 52 cards contains (i) all Kings, (ii) 3 Kings, and (iii) at least 3 Kings. (5 marks)
Ans: Total possible hands: 52C7. Veena spends her vacations in a random order visiting 4 cities (A, B, C, and D). What is the probability that she will go to (i) A before B? (ii) A before B and B before C
The number of arrangements (orders) Veena can make to visit 4 cities A, B, and C.
The sum of B, C, and D is 4, or 24.
As a result, n (S) = 24.
Because there are 24 elements in the experiment's sample space, all of these outcomes are considered equally likely. The sample space for the experiment is,
S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BDAC, BDCA, BCAD, BCDA
CABD, CADB, CBDA, CBAD, CDAB, CDBA,
DABC, DACB, DBCA, DBAC, DCAB, DCBA}
(i) Let the event ‘she visits A before B’ be denoted by E
Hence ,E = {ABCD, CABD, DABC, ABDC, CADB, DACB
ACBD, ACDB, ADBC, CDAB, DCAB, ADCB}
Therefore,
\(P (E) = \frac{n (E)}{n (S)} = \frac{12}{24} = \frac{1}{2}\)(ii) Let the event ‘Veena visits A before B and B before C’ be denoted by F.
Here F = {ABCD, DABC, ABDC, ADBC}
Thus,\(P (F) = \frac{n (F)}{n (S)} = \frac{4}{24} = \frac{1}{6}\)
Ques. Assume that 3 bulbs are chosen at random from a batch. Each bulb is examined and categorized as defective (D) or non-defective (ND) (N). What is the experiment's sample space? (2 marks)
Ans: Three bulbs will be chosen at random from the lot.
Each bulb in the batch is examined and classified as defective (D) or non-defective (ND) (N). The sample space is, S= (DDD, DDN, DND, DNN, NDD, NON, NND, NNN).
Ques. There is a coin toss. A die is thrown if the outcome is a head. The die is thrown again if the result is an even number. What is the experiment's sample space? (2 marks)
Ans: Head (H) and tail (T) are the two possible outcomes of a coin toss (T). The possible outcomes of a die throwing are 1, 2, 3, 4, 5, and 6.
Therefore, the sample space of experiment is = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65).
Ques. Assume a company has 10000 employees, of which 1000 are required to complete onsite work. All of their names are now in the basket, and 100 will be chosen at random. In this case, each employee has an equal chance of being selected? (3 marks)
Ans: Given the sample size and population, one can easily pick the probability from this dataset.
Here's the equation: The possibility of a one-time selection is as follows:
P = n/N = 1000/10000 = 10 %
Also, more than once:
P = 1-(1-(1/N))n
P = 1 – (9999/10000)n
P = 0.9
P = 9.0 %
Ques. What are the steps involved in the random sampling method? (3 marks)
Ans: The steps involved in the random sampling method are as follows:
- First, make a list of all the samples in the entire population.
- Assign a sequence number to each of the required samples.
- Determine the size of the sample.
- Compare each sample with respect to sample size.
- Lastly, use a random number generator to determine the sample using steps second and fourth.
Ques. What are the disadvantages of random sampling? (3 marks)
Ans. The disadvantages of random sampling are as follows:
- If the demographic information is incomplete, then certain groups are left.
- Random sampling means all information about the sample is not present.
- It is a time-intensive process.
- It is based on the luck or chance.
Ques. Suppose there are 6 green 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red? (2 marks)
Ans.Given: Green balls: 6
- Red Balls: 7
- Required probability = P (G) × P (R)
- (6/12) x (7/11)
- 7/22
Ques. Suppose two pair of dice are thrown together. Determine the probability that the number obtained on one face of the dice is multiple of the number obtained on the other face of the dice? (3 marks)
Ans.Given two face of dice are thrown so total number of sample formed = 62 = 36
- Since the number on a face of a die should be multiple of the other number.
- The number of possible sample are a follows:
- (1, 1) (2, 2) (3, 3) ------ (6, 6) = 6 total ways
- (2, 1) (1, 2) (1, 4) (4, 1) (1, 3) (3, 1) (1, 5) (5, 1) (6, 1) (1, 6) = 10 total ways
- (2, 4) (4, 2) (2, 6) (6, 2) (3, 6) (6, 3) = 6 total ways
- Total favorable cases formed are as follows: 6 + 10 + 6 = 22.
- So, P = 22/36
- 11/18
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