1.1 Probability Distributions
We had introduced the concepts of a probability function and random variables in our last editorial. Probability distributions describe what we think the probability of each outcome is, which is sometimes more interesting to know than simply which single outcome is most likely. They come in many shapes, but in only one size: probabilities in a distribution always add up to 1. In this section, we’ll look at a couple of popular Probability distributions.
The Bernouli Distribution
We had discussed coin tossing example in our earlier editorials. Such experiments follow the Bernoulli distribution, over two discrete outcomes—tails or heads. Think of it, however, as a distribution over 0 and 1, over 0 heads (i.e. tails) or 1 heads. Above, both outcomes were equally likely. The Bernoulli PDF (Probability Density Function) has two lines of equal height, representing the two equally-probable outcomes of 0 and 1 at either end.
The Bernoulli distribution could represent outcomes that aren’t equally likely, like the result of an unfair coin toss. Then, the probability of heads is not 0.5, but some other value p, and the probability of tails is 1-p. Like many distributions, it’s actually a family of distributions defined by parameters, like p here. When you think “Bernoulli,” just think “(possibly unfair) coin toss.”
The Binomial Distribution
The binomial distribution may be thought of as the sum of outcomes of things that follow a Bernoulli distribution. Toss a fair coin 20 times; how many times does it come up heads? This count is an outcome that follows the binomial distribution. Its parameters are n, the number of trials, and p, the probability of a “success” (here: heads, or 1). Each flip is a Bernoulli-distributed outcome, or trial. Reach for the binomial distribution when counting the number of successes in things that act like a coin flip, where each flip is independent and has the same probability of success.
Or, imagine an urn with equal numbers of white and black balls. Close your eyes and draw a ball and note whether it is black, then put it back. Repeat. How many times did you draw a black ball? This count also follows a binomial distribution.
Thus, mathematically speaking, we are interested in occurrence of an Event E exactly r times in n trials. If the probability of occurrence of the event is p, then the probability of the event occurring exactly r times in n trials is denoted by \(C(n,r)p^r(1-p)^{n-r}\), let’s have \(1-p=q\). In our next chapter, Binomial Theorem, we’ll see that this is actually the rth coefficient in the expansion \((p+q)^r\). The Expectation of the binomial distribution is \(np\).
\(P(r\ge k)=\sum\limits_{r=k}^{n}C(n,r)p^rq^{n-r}\).
Test Your Concepts:
1. For a random variable X, if the probability function is \(f(x)\), calculate \(\sum\limits_xf(x)\).
2. What is the probability that in 10 tosses of an unbiased coin, there will be at max 6 heads?
1.2 Probability computations in Uncountable Sample Spaces
Often, we may need to deal with sample spaces that are not countable in their original form. In these cases, one way is to modularise them into Geometrical structures, so that we can deal with areas and boundaries. The ‘favourable area’ for our event, compared with the area of the total sample space, can thus be used as our probability computation ratio.
Ex -1
2 persons A and B agree to meet at a common place between 11 and 12 noon. The first person to arrive shall wait for 20 mins, and then leaves. Assuming that their times of arrival are independent and random, what is the probability that A and B meet?
Solution:
If A and B arrive at \(x,y\) minutes after 11, then for the condition to meet, we have \(|x-y|\le 20\).
Consider the below figure:
Fig 1.1
If we consider the square \(OABC\), then we’re basically modularising the entire sample space to a geometrical space. While, the ‘favourable area’ happens to be the area of the region \(OSRBQP\).
This area is basically = \(OABC-\triangle SCR-\triangle APQ=3600-\frac{1}{2}(40)(40)-\frac{1}{2}(40)(40)=2000\text{ sq. units}\).
The total area of the square = \(60\times60=3600\text{ sq. units}\).
Thus probability of the meeting = \(\frac{2000}{3600}=\frac{5}{9}\).
Ex -2
Let \(\mathbb{X}\) denote the set of points in the \(x,y\) plane where both the points are integers. If a coin of diameter \(\frac{1}{2}\) is randomly placed on the plane, find the probability that the coin covers point of \(\mathbb{X}\).
Solution:
Fig 1.2
Consider the above figure. Let \(S\) denote the set of points inside a square with corners \((a,b),(a,b+1),(a+1,b),(a+1,b+1)\in\mathbb{X}\). Also, let \(P\) denote the set of points with distance less than \(\frac{1}{4}\) from the corner area. Thus area of \(P=\pi \times \left(\frac{1}{4}\right)^2\). Area of \(S=1\), thus probability is \(\frac{\text{Area of P}}{\text{Area of S}}\frac{\pi}{16}\).