# Traffic management

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### 3.2.1 The Binomial Distribution

The situation often used to introduce the binomial distribution is that of n independent trials or experiments, each of which can have only one of two possible outcomes, often designated as ‘success’ and ‘failure’. The probabilities of the outcomes known are the same for each trial and, because only the two outcomes are possible, necessarily add to one. A simple example of such a situation is a series of n tosses of a true coin, in which only the two outcomes ‘heads’ and ‘tails’ are possible and each has a probability of 0.5.

A discrete probability distribution is appropriate to predict the number of ‘successes’ obtained in n trials or the number of ‘heads’ in n tosses of the coin.

The binomial frequency distribution can be written as:

 $\mathrm{b(x) = }\left( \frac{\mathrm{n}}{\mathrm{x}} \right)\mathrm{p}^{\mathrm{x}}{\mathrm{(1} - \mathrm{p)}}^{\mathrm{n - x}}$ 3.1 where b(x) = the probability of exactly x successes in n trials (x ≤ n) p = the probability of a success in any one trial $\left( \frac{\mathrm{n}}{\mathrm{x}} \right)$ = the binomial coefficient, equal to the number of different combinations of x items that can be formed from a group of n items $$\frac{\mathrm{n!}}{\mathrm{x!}\left( \mathrm{n} - \mathrm{x} \right)\mathrm{!}}$$

The mean of the binomial frequency distribution, or the expected number of ‘successes’ in n trials if p is the probability of a ‘success’ in one trial, is:

 $E(\text{ x })= np$ 3.2

And the variance of x is:

 $\sigma ^{2}(\text{ x })= n p(1-p)$ 3.3

In traffic applications, the cumulative form of the binomial distribution is often useful. This can be written as:

 $B (X) = \sum _{\text{ x }=0}^{X}b (\text{ x })$ 3.4 where B(x) the probability of X or less successes in n trials b(x) is as defined in Equation 3.1

#### Example application

The situation of vehicles entering a car park through a particular gate (say Gate A) provides an example of how the binomial distribution might be applied in a traffic context. Assume that, at a certain time of day, the car park is being accessed by 300 veh/h and that 35% of these, on average, use Gate A. Assume also that the aim is to determine the probability of more than six vehicles using Gate A over a two minute period.

On average, 10 vehicles will access the car park over a two minute period. These can be considered as 10 independent trials, in each of which there is a probability of 0.35 that the vehicle will use Gate A and a probability of 0.65 that it will use another gate. The probability that more than six of the 10 vehicles will use Gate A is 1 – B(6) where B(6) is calculated using Equation 3.4 with X = 6 and each b(x). x = 1,..., 6, calculated from Equation 3.1, with n = 10 and p = 0.35. It is found that there is a probability of 0.0260, or 2.6%, that more than six vehicles will use Gate A.

Note that the same answer can be obtained as the probability of three or less of the 10 vehicles using another gate, rather than Gate A, that is, B(3), calculated using Equation 3.4 with X = 3 and each b(x), x = 1,..., 3, being calculated from Equation 3.1, with n = 10 and p = 0.65.