Let $X$ denote the number of hours you play during a randomly selected day. The probability that $X$ can take values $x$ has the following form, where $c$ is some constant.

$\mathrm{P}(\mathrm{X}=\mathrm{x})=\left\{\begin{array}{lll} 0.1, & \text { if } \mathrm{x}=0 \\ \mathrm{cx}, & \text { if } \mathrm{x}=1 \text { or } \mathrm{x}=2 \\ \mathrm{c}(5-\mathrm{x}), & \text { if } \mathrm{x}=3 \text { or } \mathrm{x}=4 \\ 0, & \text { otherwise } \end{array}\right.$

Match List-I with List-II :

List-I	List-II
(A) $c$	(I) 0.75
(B) $P(X \leq 2)$	(II) 0.3
(C) $P(X = 2)$	(III) 0.55
(D) $P(X \geq 2)$	(IV) 0.15

Let's find the value of constant $c$.

Since this is a probability distribution, the sum of all probabilities must equal 1.

$P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1$

$0.1 + c(1) + c(2) + c(5-3) + c(5-4) = 1$

$0.1 + c + 2c + 2c + c = 1$

$0.1 + 6c = 1$

$6c = 0.9$

$c = 0.15$

---

Now let's calculate each probability:

$P(X = 0) = 0.1$

$P(X = 1) = c(1) = 0.15$

$P(X = 2) = c(2) = 0.3$

$P(X = 3) = c(5-3) = 0.3$

$P(X = 4) = c(5-4) = 0.15$

---

Let's find the requested probabilities:

(A) $c = 0.15$

(B) $P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1 + 0.15 + 0.3 = 0.55$

(C) $P(X = 2) = 0.3$

(D) $P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.3 + 0.3 + 0.15 = 0.75$

---

Therefore, the matching is:

(A) $c$ → (IV) 0.15

(B) $P(X \leq 2)$ → (III) 0.55

(C) $P(X = 2)$ → (II) 0.3

(D) $P(X \geq 2)$ → (I) 0.75