Probability Distributions

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RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

Suppose that a random experiment consists of tossing two dice, and the quantity of interest is the sum of the numbers on the two dice. Let us denote this sum by X.

X will be termed a random variable of this experiment and it can take one of these possible values: 2, 3, 4, ...., 12.

Consider another random experiment wherein we toss a coin a 100 times and we are interested in the number of Heads obtained, which is again a random variable of this experiment, with one of these possible values: 0, 1, 2, ...... 100

Of course, an experiment can have many random variables associated with it. For example, for the coin tossing experiment above, we can have so many possible random variables.

\[\begin{align}& {{X}_{1}}:\text{ }No.\text{ }of\text{ }tails \\ & {{X}_{2}}:\text{ }No.\text{ }of\text{ }Heads-\text{ }\text{ }No.\text{ }of\text{ }tails \\ & {{X}_{3}}:\text{ }The\text{ }no.\text{ }of\text{ }tosses\text{ }in\text{ }the\text{ }largest\text{ }consecutive\text{ }sequence\text{ }of\text{ }Heads \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \vdots  \\&\qquad\qquad\qquad\qquad\qquad\qquad\qquad etc \\ \end{align}\]

Thus, we see that a random variable is a way of assigning to each outcome of the experiment, a single real number, which will vary with different outcomes of the experiment. So far, so good.

Now, we will try to understand what the probability distribution of a random variable means.

Consider once again the random experiment of rolling two dice and observing the sum of the numbers on the two dice, which we denoted by X. X can take a multitude of values in the following ways:

X

Outcome(s) which gives this X

2

3

4

5

6

7

8

9

10

11

12

(1, 1)

(1, 2) (2, 1)

(1, 3) (2, 2 ) (3, 1)

(1, 4) (2, 3) (3, 2) (4, 1)

(1, 5) (2, 4) (3, 3) (4, 2) (5, 1)

(1, 6) (2, 5) (3, 4) (4, 3) (5, 2) (6, 1)

(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)

(3, 6) (4, 5) (5, 4) (6, 3)

(4, 6) (5, 5) (6, 4)

(5, 6) (6, 5)

(6, 6)

Each value of X has a certain probability of being obtained. For example

\[\begin{align}& P(X=5)=\frac{\text{No}\text{.}\ \text{of outcomes for which }X\text{ = 5}}{\text{Total No}\text{. of outcomes}} \\ & \qquad\qquad=\frac{4}{36} \\ & \qquad\qquad =\frac{1}{9} \\ \end{align}\]

Let us plot the probabilities for each value of X:

X 2 3 4 5 6 7 8 9 10 11 12
  P(X) \(\begin{align}\frac{1}{36}\end{align}\) \(\begin{align}\frac{1}{18}\end{align}\) \(\begin{align}\frac{1}{12}\end{align} \) \(\begin{align}\frac{1}{9}\end{align}\) \(\begin{align}\frac{5}{36}\end{align}\) \(\begin{align}\frac{1}{6}\end{align}\) \(\begin{align}\frac{5}{36}\end{align}\) \(\begin{align}\frac{1}{9}\end{align}\) \(\begin{align}\frac{1}{12}\end{align}\)     \(\begin{align}\frac{1}{18}\end{align}\) \(\begin{align}\frac{1}{36}\end{align} \)

This table gives us what is known as the probability distribution (PD) of X, that is, it is a description of how the “probability is distributed” across different values of the random variable. In simple words, the PD of any random variable (RV) X tells us how probable each value of the RV is.

The sum of the various probabilities in a PD must be 1, as should be obvious. This fact you are urged to confirm for the last table.

Let us write down another PD as an example.

Example – 17

Find the PD of the number of doublets in three throws of a pair of dice.

Solution: In three throws, we can obtain either 0, 1, 2, or 3 doublets. Denote this RV by X.

Now, doublets can be obtained in 6 ways, namely (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6), out of a total possible ways of 6 × 6 = 36. Thus, the probability of obtaining a doublet is

\[\begin{align} & \qquad\quad{{p}_{1}}=P(\text{Doublet})=\frac{1}{6} \\ &\text{and } \quad{{p}_{2}}=P(\text{Not a Doublet})=\frac{5}{6} \\ \end{align}\]

Now, let us obtain the probabilities for the different values of X.

\[\begin{align}
  & X=0\Rightarrow P(X=0)=P\left\{ \text{No doublets in all three throws} \right\} \\ 
 &\qquad\qquad\qquad\qquad\; =\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6} \\ 
 &\qquad\qquad\qquad\qquad\;  =\frac{125}{216} \\ 
 & X=1\Rightarrow P(X=1)=P\left\{\text{  A doublet exactly once} \right\} \\ 
 & \qquad\qquad\qquad\qquad\; =\left\{ \frac{1}{6}\times \frac{5}{6}\times \frac{5}{6} \right\}\times 3\,\,\left( \text{why is there the factor of 3?} \right) \\ 
 &\qquad\qquad\qquad\qquad\;  =\frac{75}{216} \\ 
 & X=2\Rightarrow P(X=2)=P\left\{ \text{ Exactly two doublets} \right\} \\ 
 & \qquad\qquad\qquad\qquad\; =\left\{ \frac{1}{6}\times \frac{1}{6}\times \frac{5}{6} \right\}\times 3\,\,\,\left( \text{ why a factor of 3 again?} \right) \\ 
 &\qquad\qquad\qquad\qquad\;  =\frac{15}{216} \\ 
 & X=3\Rightarrow P(X=3)=P\left\{ \text{ Doublets in all three throws} \right\} \\ 
 &\qquad\qquad\qquad\qquad\;  =\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6} \\ 
 &\qquad\qquad\qquad\qquad\;  =\frac{1}{216} \\ 
\end{align}\]

Thus, the PD is

 X 0 1 2 3
 P(X) \(\begin{align}\frac{125}{216}\end{align}\) \(\begin{align}\frac{75}{216}\end{align}\) \(\begin{align}\frac{15}{216}\end{align}\) \(\begin{align}\frac{1}{216}\end{align}\)

Note the (obvious) fact that \(\sum\limits_{i=0}^{3}{P(X=i)=1}\)

Example – 18

Find the PD of the number of Heads in four tosses of a coin.

Solution: Let X denote the number of Heads in the four tosses

Verify the following probabilities:

\[\begin{align}
  & P(X=0)=\frac{1}{{{2}^{4}}}=\frac{1}{16} \\ 
 & P(X=1)={{\,}^{4}}{{C}_{1}}\times \frac{1}{{{2}^{4}}}=\frac{1}{4} \\ 
 & P(X=2)={{\,}^{4}}{{C}_{2}}\times \frac{1}{{{2}^{4}}}=\frac{3}{8} \\ 
 & P(X=3)={{\,}^{4}}{{C}_{3}}\times \frac{1}{{{2}^{4}}}=\frac{1}{4} \\ 
 & P(X=4)=\,\frac{1}{{{2}^{4}}}=\frac{1}{16} \\ 
\end{align}\]

Thus, the PD is

X 0 1  2
 P(X) \(\begin{align}\frac{1}{16}\end{align}\)  \(\begin{align}\frac{1}{4}\end{align}\) \(\begin{align}\frac{3}{8}\end{align}\)   \(\begin{align}\frac{1}{4}\end{align}\) \(\begin{align}\frac{1}{16}\end{align}\)