Probability Mass Function (p. m. f.)

Sometimes the probability $p_i$ of $X$ taking the value $x_i$ can be written as an explicit function of $x_i$, valid for every possible value of $X$; such a function is called the probability mass function (p.m.f.) of the discrete random variable $X$. A first illustration is the coin-tossing experiment where $X$ is the number of tosses needed to get the first head. If the probability of a head on a single toss is $t$ (so the probability of a tail is $1 - t$), then $X$ can equal any positive integer and $P[X = i] = (1-t)^{i-1} t$ for $i = 1, 2, 3, \ldots$. This is easy to justify: getting the first head on toss $i$ means the first $i-1$ tosses were all tails (probability $(1-t)^{i-1}$) followed by a head on toss $i$ (probability $t$).

Formally: let the possible values of a discrete random variable $X$ be $x_1, x_2, x_3, \ldots$, with $p_i = P[X = x_i]$ for $i = 1, 2, \ldots$. The function $p$ is called the probability mass function of $X$ if (i) $p_i \ge 0$ for every $i$, and (ii) $\sum_{i=1}^n p_i = 1$. As a concrete illustration, toss a fair coin 4 times and let $X$ be the number of heads; $X$ can be 0, 1, 2, 3 or 4, and its p.m.f. is exactly Table 7.3. This distribution follows the closed-form rule $P[X = x] = {}^4C_x \left(\dfrac{1}{2}\right)^4$, $x = 0, 1, 2, 3, 4$, where ${}^4C_x$ counts the number of ways of getting $x$ heads out of 4 tosses — the same counting idea used throughout the worked examples below.

Worked examples.

Example 1. Two persons A and B toss a coin thrice; each head gives A ₹2 from B, each tail gives B ₹1.5 from A. Let $X$ be the net amount gained (or lost) by A. Show $X$ is a discrete random variable and describe it as a function on the sample space.

Since $X$ is a number that depends on the outcome of a random experiment, it is a random variable; because the sample space has only $2^3 = 8$ outcomes, $X$ is discrete. With $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$, the value of $X$ in rupees for each outcome is: $X(HHH) = 2 \times 3 = ₹6$; $X(HHT) = X(HTH) = X(THH) = 2(2) - 1.5(1) = ₹2.50$; $X(HTT) = X(THT) = X(TTH) = 2(1) - 1.5(2) = -₹1.00$; $X(TTT) = -1.5 \times 3 = -₹4.50$ (a negative value means a loss to A). Every outcome maps to a unique value of $X$, confirming $X$ is a function on $S$, with possible values $-4.50, -1, 2.50, 6$.

Example 2. A bag has 1 red and 2 green balls ($r, g_1, g_2$). A ball is drawn, its colour noted, and replaced; a second ball is then drawn and noted. Let $X$ be the number of red balls drawn. Derive the probability distribution of $X$.

The sample space (9 equally likely outcomes) is $S = \{rr, rg_1, rg_2, g_1r, g_2r, g_1g_1, g_1g_2, g_2g_1, g_2g_2\}$. Here $X(rr) = 2$; $X(rg_1) = X(rg_2) = X(g_1r) = X(g_2r) = 1$ (4 outcomes); $X(g_1g_1) = X(g_1g_2) = X(g_2g_1) = X(g_2g_2) = 0$ (4 outcomes). So $X \in \{0, 1, 2\}$ with $P[X=0] = 4/9$, $P[X=1] = 4/9$, $P[X=2] = 1/9$.

Example 3. Two cards are drawn with replacement from a well-shuffled deck of 52. Find the probability distribution of the number of aces drawn.

Since the draws are independent (with replacement), $P[\text{ace}] = 4/52 = 1/13$ and $P[\text{non-ace}] = 12/13$. With $X$ the number of aces among the two draws, $X \in \{0, 1, 2\}$: $P[X=0] = (12/13)^2 = 144/169$; $P[X=1] = 2 \times (1/13)(12/13) = 24/169$; $P[X=2] = (1/13)^2 = 1/169$.

Example 4. A fair die is thrown; $X$ is the number of factors of the number shown on top. Find the probability distribution of $X$.

The number of positive factors of 1,2,3,4,5,6 are 1,2,2,3,2,4 respectively, so $X(1)=1$, $X(2)=X(3)=X(5)=2$, $X(4)=3$, $X(6)=4$. Hence $p_1 = P[X=1] = P\{1\} = 1/6$; $p_2 = P[X=2] = P\{2,3,5\} = 3/6$; $p_3 = P[X=3] = P\{4\} = 1/6$; $p_4 = P[X=4] = P\{6\} = 1/6$.

Example 5. Find the probability distribution of the number of doublets in three throws of a pair of dice. …