Q.Two fair coins are tossed together, and $X$ denotes the number of heads obtained. Construct the probability distribution (probability mass function) of $X$, and verify that it satisfies the conditions of a valid probability distribution.

Since two coins are tossed, the four equally likely outcomes are HH, HT, TH, TT, and $X$ (the number of heads) can be 0, 1, or 2 depending on the outcome.

The sample space when two fair coins are tossed is $S = \{HH, HT, TH, TT\}$, with each of the 4 outcomes equally likely, so each has probability $\tfrac{1}{4}$.

Let $X$ = number of heads obtained:

• $TT$ gives $X = 0$
• $HT$ and $TH$ each give $X = 1$
• $HH$ gives $X = 2$

So $X$ can take the values $0, 1, 2$, with:

This gives the probability distribution table:

Verification of validity:

• Every $p_i \geq 0$: yes, $\tfrac{1}{4}, \tfrac{1}{2}, \tfrac{1}{4}$ are all positive.
• $\sum p_i = 1$: $\tfrac{1}{4} + \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{1}{4} + \tfrac{2}{4} + \tfrac{1}{4} = \tfrac{4}{4} = 1$, so this condition also holds.

Both conditions for a valid p.m.f. are satisfied.

Alternatively, since $X$ = number of heads in 2 tosses follows the Binomial pattern with $n=2$ and $p=\tfrac{1}{2}$ for each coin, $P(X=k) = \binom{2}{k}\left(\tfrac{1}{2}\right)^k\left(\tfrac{1}{2}\right)^{2-k}$, which gives the same values: $P(X=0)=\tfrac{1}{4}$, $P(X=1)=\tfrac{1}{2}$, $P(X=2)=\tfrac{1}{4}$.

A frequent error is treating HT and TH as a single outcome instead of two separate equally likely outcomes, which wrongly gives $P(X=1) = \tfrac{1}{3}$ instead of $\tfrac{1}{2}$ and breaks the probabilities' sum. Always list every outcome of the sample space separately before grouping them by the value of $X$.