Even Numbered Vehicles in the Odd Numbered Vehicle Day and the Theoretical Two-Category Discrete Probability Distribution Without Replacement, Statistical Note 33

Traffic Police Office of a X Metropolitan City in Y country announced odd-even rule on alternate days for vehicles. This rule means that general vehicles (non-emergency) with number plates ending in odd numbers will be allowed to ply the roads on odd calendar days whereas those ending in even numbers will be permitted on even days. It was estimated that on the first odd day of the rule, two percent of the vehicles plied the road were even numbered either due to ignorance or violation of the rule. A sample of 20 vehicles was taken on that first odd day to see how many even numbered vehicles plied on the odd day. Calculate the theoretical discrete probability distribution of even numbered vehicles.

This question popped up to me when Kathmandu Metropolitan Traffic Police Division imposed odd-even rule on alternate days for vehicles in Kathmandu valley during the fourth Bay of Bengal Initiative for Multi-Sectoral Technical and Economic Cooperation summit held on August 30 and 31.

This is an example of the two-category discrete probability distribution of sampling without replacement. Refer to my earlier Statistical Notes for clarity on calculating the two-category discrete probability of samples drawn without replacement using tree diagram, formula and Excel software function.

Theoretical Discrete Probability Distribution Without Replacement

This example has three characteristic features. First, the example has a finite population of one million vehicles plied on the road on the odd day, denoted by ‘N’. There are ‘M’ even numbered vehicles and ‘N-M’ odd numbered vehicles on that day. Second, each vehicle can be characterized as a success or a failure ‘Odd’. Since the question asks the probability of even numbered, the selection of a ‘Even numbered vehicle’ is considered as a success, and there are one million even numbered vehicles in the population. Third, a sample of 20 vehicles, denoted by ‘n’, is sampled without replacement in a way that each sample of 20 vehicles is equally likely to be selected.

Let X be a random variable of interest that takes one of 0 to  20 values as the number of even numbered vehicle in the sample of 20 vehicles sampled without replacement, denoted by ‘x’. The probability distribution of X depends on the parameters, ‘n’, ‘M’ and ‘N’, and is given by the expression

P(X=x) = h(x;n,M,N) = Number of outcomes having X=x divided by total number of outcomes

P(X=x) = h(x;n,M,N) = [C(M,x) X C(N-M,n-x)]/C(N,n)

This distribution is referred to as Hypergeometric distribution.

In this example, n=20, N=1,000,000, M is two percent of N, equal to 20,000,  and ‘x’ takes the value 0 to 20. Putting these values in the above formula, one gets for example:

P(X=0) = [C(20000,0) X C(980000,20)/C(1000000,20)] = 0.667605383

This value is equal to the one presented in table 1 using formula manually and Excel Hypergeometric function. In the same way, the probability for other number of even numbered vehicles in 20 vehicles sampled without replacement can be calculated.

Table 1: Number of Success (Even Numbered Vehicle), Failure (Odd Numbered Vehicle) in a Sample of 20 Vehicles on the Odd Numbered Vehicle Day and Probability Distribution Without Replacement

Graphical Presentation

Chart 1 shows the theoretical two category discrete probability distribution of even numbered vehicles in 20 vehicles sampled without replacement from the population, the total number of vehicles that plied the road on odd vehicle day.  This clearly shows the highly skewed curve of the theoretical probability distribution in population that has only two percent even numbered vehicles that plied the road on the odd vehicle day.

Conclusion

Looking at the table and the chart one can see that no even numbered vehicle or all odd numbered vehicle on the odd numbered vehicle day in a sample of 20 vehicles sampled without replacement is highly likely. The likelihood decreases with the increase in the number of even numbered vehicles in a sample of 20 vehicles and becomes unlikely after increase in the number of even numbered vehicles from 11 to all 20 even numbered vehicles in a sample of 20 vehicles sampled without replacement.