Maths > Probability > 12.0 Probability Distribution
Probability
1.0 Basic Definitions
2.0 Basic Notations
3.0 Probability
4.0 Intersection and Union of Sets of Events
5.0 Conditional Probability
6.0 Multiplication Theorem
7.0 Independent Events
7.1 Occurrence of at least one of the independent events
7.2 Pairwise independent events
7.3 Mutually independent events
8.0 Total Probability Theorem
9.0 Bayes' Theorem
10.0 Illustration for understanding the difference between total probability theorem and baye's theorem
11.0 Probability Distribution of Random Variables
12.0 Probability Distribution
13.0 Mean and variance of a discrete random variable
14.0 Binomial Distribution for Successive Events
15.0 Mean and variance of binomial distribution
12.1 Basic Equations
7.2 Pairwise independent events
7.3 Mutually independent events
If $X$ is a random with probability distribution
| $X$: | ${x_1}$ | ${x_2}$ | ${x_3}$ | ...... | ${x_n}$ |
| $P(X)$: | ${p_1}$ | ${p_2}$ | ${p_3}$ | ...... | ${p_n}$ |
Then,
$$\begin{equation} \begin{aligned} i.\;\;P(X \le {x_i}) = P(X = {x_1}) + P(X = {x_2}) + ... + P(X = {x_i}) = {p_1} + {p_2} + .... + {p_i} \\ ii.\;\;P(X < {x_i}) = P(X = {x_1}) + P(X = {x_2}) + ... + P(X = {x_{i - 1}}) = {p_1} + {p_2} + .... + {p_{i - 1}} \\ iii.\;\;P(X \ge {x_i}) = P(X = {x_i}) + P(X = {x_{i + 1}}) + ... + P(X = {x_n}) = {p_i} + {p_{i + 1}} + .... + {p_n} \\ iv.\;\;P(X > {x_i}) = P(X = {x_{i + 1}}) + P(X = {x_{i + 2}}) + ... + P(X = {x_n}) = {p_{i + 1}} + {p_{i + 2}} + .... + {p_n} \\ v.\;\;P(X \le {x_i}) = 1 - P(X > {x_i}) \\ vi.\;\;P(X < {x_i}) = 1 - P(X \ge {x_i}) \\ vii.\;\;P(X \ge {x_i}) = 1 - P(X < {x_i}) \\ viii.\;\;P(X > {x_i}) = 1 - P(X \le {x_i}) \\\end{aligned} \end{equation} $$
Illustration 41. Determine which of the following can be probability distributions of a random variable $X$:
i.
| $X$: | $0$ | $1$ | $2$ |
| $P(X)$: | $0.4$ | $0.4$ | $0.2$ |
ii.
| $X$: | $0$ | $1$ | $2$ |
| $P(X)$: | $0.6$ | $0.1$ | $0.2$ |
Solution:
i. Clearly,
$$\begin{equation} \begin{aligned} P(X = 0) + P(X = 1) + P(X = 2) \\ = 0.4 + 0.4 + 0.2 \\ = 1 \\\end{aligned} \end{equation} $$
Hence the given distribution of probabilities is a probability distribution of random variable $X$.
ii. Here,
$$\begin{equation} \begin{aligned} P(X = 0) + P(X = 1) + P(X = 2) \\ = 0.6 + 0.1 + 0.2 \\ = 0.9 \\\end{aligned} \end{equation} $$
Thus,
$$P(X = 0) + P(X = 1) + P(X = 2) \ne 1$$
Hence the given distribution of probabilities is not a probability distribution.
Illustration 42. The random variable $X$ has a probability distribution $P(X)$ of the following form, where $k$ is some number: $$P(X = x)\left\{ \matrix{k,\;\;if\;x = 0 \hspace{1em} \cr 2k,\;\;if\;x = 1 \hspace{1em} \cr 3k,\;\;if\;x = 2 \hspace{1em} \cr 0,\;\;otherwise \hspace{1em} \cr} \right.$$
i. Determine the value of $k$
ii. Find $P(X < 2),P(X \le 2),P(X \ge 2)$
Solution: Let us chart the random variable and the corresponding probability distribution.
| $X$ | $0$ | $1$ | $2$ |
| $P(X)$ | $k$ | $2k$ | $3k$ |
i. As per the necessary conditions, $\sum\limits_{i = 1}^n {{p_i}} = 1$,
$$\begin{equation} \begin{aligned} P(X = 0) + P(X = 1) + P(X = 2) = 1 \\ \Rightarrow k + 2k + 3k = 1 \\ \Rightarrow 6k = 1 \\ \Rightarrow k = {1 \over 6} \\\end{aligned} \end{equation} $$
ii. $$\begin{equation} \begin{aligned} P(X < 2) \\ P(X < 2) = P(X = 0) + P(X = 1) \\ = k + 2k = 3k = 3\left( {{1 \over 6}} \right) = {1 \over 2} \\ P(X \le 2) \\ P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2) \\ = k + 2k + 3k = 6\left( {{1 \over 6}} \right) = 1 \\ P(X \ge 2) \\ P(X \ge 2) = P(X = 2) + P(X = 3) + P(X = 4) + .... \\ = 1 - P(X < 2) = 1 - {1 \over 2} = {1 \over 2} \\\end{aligned} \end{equation} $$
Illustration 43. Four bad oranges are accidentally mixed with $16$ good oranges. Find the probability distribution of the number of bad oranges in a draw of two oranges.
Solution: Let $X$ denote the number of bad oranges in a draw of $2$ oranges drawn from group of $16$ good oranges and $4$ bad oranges.
The possibilities are,
i. none of them is bad.
i.e. $X=0$
ii. one of them is bad.
i.e. $X=1$
iii. both of them are bad.
i.e. $X=2$
Therefore $X$ can take the values $0$, $1$ and $2$.
Now,
$$\begin{equation} \begin{aligned} P(X = 0) = probability\;of\;getting\;no\;bad\;orange \\ = probability\;of\;getting\;two\;good\;oranges \\ P(X = 0) = {{^{16}{C_2}} \over {^{20}{C_2}}} = {{12} \over {19}} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} P(X = 1) = probability\;of\;getting\;one\;bad\;orange \\ P(X = 1) = {{^{16}{C_1} \times {\;^4}{C_1}} \over {^{20}{C_2}}} = {{32} \over {95}} \\\end{aligned} \end{equation} $$
$$\begin{equation} \begin{aligned} P(X = 2) = probability\;of\;getting\;two\;bad\;oranges \\ P(X = 2) = {{^4{C_2}} \over {^{20}{C_2}}} = {3 \over {95}} \\\end{aligned} \end{equation} $$
Thus the probability distribution of $X$ is given by,
| $X$ | $0$ | $1$ | $2$ |
| $P(X)$ | ${{12} \over {19}}$ | ${{32} \over {95}}$ | ${3 \over {95}}$ |
Illustration 44. Find the probability distribution of the number of green balls drawn when $3$ balls are drawn, one by one, without replacement from a bag containing $3$ green and $5$ white balls.
Solution: Let $X$ denote the total number of green balls drawn in three draws without replacement.
Let ${G_i}$ be the event of getting a green ball in the ${i^{th}}$ draw.
The possibilities are
i. None are green
i.e. $X=0$
ii. One is green
i.e. $X=1$
iii. Two are green
i.e. $X=2$
iv. All the three are green
i.e. $X=3$
$X=0$
$$\begin{equation} \begin{aligned} P(X = 0) = probability\;of\;getting\;no\;green\;ball\;in\;three\;draws \\ = P({{\bar G}_1} \cap {{\bar G}_2} \cap {{\bar G}_3})\; \\ i.e.\;getting\;white\;ball\;in\;all\;three\;draws \\ = P({{\bar G}_1})\;P({{\bar G}_2}/{{\bar G}_1})\;P({{\bar G}_3}/{{\bar G}_2} \cap {{\bar G}_1}) \\ \;Since,\;the\;drawing\;of\;the\;balls\;obeys\;conditional\;probability. \\ = {5 \over 8} \times {4 \over 7} \times {3 \over 6} = {5 \over {28}} \\\end{aligned} \end{equation} $$
$X=1$
$$\begin{equation} \begin{aligned} P(X = 1) = probability\;of\;getting\;one\;green\;ball\;in\;three\;draws \\ = P(({G_1} \cap {{\bar G}_2} \cap {{\bar G}_3}) \cup ({{\bar G}_1} \cap {G_2} \cap {{\bar G}_3}) \cup ({{\bar G}_1} \cap {{\bar G}_2} \cap {G_3})) \\ = P({G_1} \cap {{\bar G}_2} \cap {{\bar G}_3}) + P({{\bar G}_1} \cap {G_2} \cap {{\bar G}_3}) + P({{\bar G}_1} \cap {{\bar G}_2} \cap {G_3}) \\ = P({G_1})P({{\bar G}_2}/{G_1})P({{\bar G}_3}/{G_1} \cap {{\bar G}_2}) + P({{\bar G}_1})P({G_2}/{{\bar G}_1})P({{\bar G}_3}/{G_2} \cap {{\bar G}_1}) + P({{\bar G}_1})P({{\bar G}_2}/{{\bar G}_1})P({G_3}/{{\bar G}_1} \cap {{\bar G}_2}) \\ = {3 \over 8} \times {5 \over 7} \times {4 \over 6} + {5 \over 8} \times {3 \over 7} \times {4 \over 6} + {5 \over 8} \times {4 \over 7} \times {3 \over 6} = {{15} \over {28}} \\\end{aligned} \end{equation} $$
$X=2$
$$\begin{equation} \begin{aligned} P(X = 2) = probability\;of\;getting\;two\;green\;balls\;in\;three\;draws \\ = P\left( {({G_1} \cap {G_2} \cap {{\bar G}_3}) \cup ({{\bar G}_1} \cap {G_2} \cap {G_3}) \cup ({G_1} \cap {{\bar G}_2} \cap {G_3})} \right) \\ = P({G_1} \cap {G_2} \cap {{\bar G}_3}) + P({{\bar G}_1} \cap {G_2} \cap {G_3}) + P({G_1} \cap {{\bar G}_2} \cap {G_3}) \\ = P({G_1})P({G_2}/{G_1})P({{\bar G}_3}/{G_1} \cap {G_2}) + P({{\bar G}_1})P({G_2}/{{\bar G}_1})P({G_3}/{G_2} \cap {{\bar G}_1}) + P({G_1})P({{\bar G}_2}/{G_1})P({G_3}/{G_1} \cap {{\bar G}_2}) \\ = {3 \over 8} \times {2 \over 7} \times {5 \over 6} + {5 \over 8} \times {3 \over 7} \times {2 \over 6} + {3 \over 8} \times {5 \over 7} \times {2 \over 6} = {{15} \over {56}} \\\end{aligned} \end{equation} $$
$X=3$
$$\begin{equation} \begin{aligned} P(X = 3) = probability\;of\;getting\;three\;green\;balls\;in\;three\;draws \\ = P({G_1} \cap {G_2} \cap {G_3}) \\ = P({G_1})P({G_2}/{G_1})P({G_3}/{G_1} \cap {G_2}) \\ = {3 \over 8} \times {2 \over 7} \times {1 \over 6} = {1 \over {56}} \\\end{aligned} \end{equation} $$
Thus, the probability distribution of the number of green balls is given by,
| $X:$ | $0$ | $1$ | $2$ | $3$ |
| $P(X):$ | ${5 \over {28}}$ | ${{15} \over {28}}$ | ${{15} \over {56}}$ | ${1 \over {56}}$ |
