Let $Y$ be a bernoulli random variable defined such that each flip
$$
Y= \{ T, H \} :== \{ 0, 1 \} \text{, where T :== Tails and H :== Heads}$$
$$P(Y = y) =\begin{cases}0.5, &y = 0 \\0.5, &y = 1 \\0, & \text{otherwise}\end{cases}$$
Let $X$ be a random variable = number of Heads (defined as 1's) in four independent flips of a fair coin, e.g. four bernoulli trials
$$X = \sum_{i=1}^{4} Y(i)
$$
$$\Rightarrow \text{ X can take on the values } \{0, 1, 2, 3, 4\} \text{ heads.}$$
To construct the pmf for $X$ let's compute the probability for possible value of $X = x$
| Number of H, X=x | P(X=x) as formula | P(X=x) as value |
|---|---|---|
| 0 | $$( \binom{4}{0} \times 0.5^0 \times 0.5^4 )$$ | 0.0625 |
| 1 | $$( \binom{4}{1} \times 0.5^1 \times 0.5^3 )$$ | 0.25 |
| 2 | $$( \binom{4}{2} \times 0.5^2 \times 0.5^2 )$$ | 0.375 |
| 3 | $$( \binom{4}{3} \times 0.5^3 \times 0.5^1 )$$ | 0.25 |
| 4 | $$( \binom{4}{4} \times 0.5^4 \times 0.5^0 )$$ | 0.0625 |
As a check on this table, if represents a legitimate pmf for $X$, we see that for all values making up the supp($X$) the probabilities are greater than 0 and the sum of the probabilities is = 1.
So, the soution to part 1 is
$$\Rightarrow X \in \{ 0, 1, 2, 3, 4 \} \text{, the pmf in bracket notation is: }$$
$$P(X=x) = \begin{cases} 0.0625, &x = 0\\ 0.25, &x = 1\\ 0.375, &x = 2\\ 0.25, &x = 3\\ 0.0625, &x = 4\\ 0, &otherwise\end{cases}$$
For part 2 ...
$$P(\text{ X is odd }) \Rightarrow P(\text{ X = 1 or X = 3 }) =\text{ (By Independence) } P(X = 1) + P( X = 3) = 0.5$$
Finally for part 3 we can compute the cdf $( F_{X}(x) )$ for $X$ by recalling that
$$( F_{X}(x) ) :== P(X \leq x)$$
That is we sum all the mass from $( -\infty )$ to desired value of $x$ and recall that this cumulative probability quantity only changes at values of $X$ for which there is positive mass. Since as we recall probability is only in the range of 0 to 1, and recalling that cdf increases at increasing supp values, we have...
$$F_{X}(x) =\begin{cases}0 & \text{if } x < 0 \\0.0625 & \text{if } 0 \leq x < 1 \\0.3125 & \text{if } 1 \leq x < 2 \\0.6875 & \text{if } 2 \leq x < 3 \\0.9375 & \text{if } 3 \leq x < 4 \\1 & \text{if } 4 \leq x \\\end{cases}$$
For part 4 we can just read off of the cdf in bracket notation
$$P(X \leq 2) \text{ is the sum of the probability mass values at 0, 1, 2 or from the bracket notation = } 0.6875$$
$$P(X \leq 1.75) \text{ = 0.3125 and is computed by recognizing that the cdf is defined for all values of } ( \mathbb{R} ) \text{, increasing at increasing supp values}$$
Finally,
$$P(X \geq 2.3) = 1 - P(X \lt 2.3) \text{ which from the bracket notation = } 1 - 0.6875 = 0.3125$$