Question

\(\text{Let } X \text{ denote the number of hours you play during a randomly selected day. The probability that } X \text{ can take values } x \text{ has the following form, where } c \text{ is some constant:}\)
\(P(X = x) = \begin{cases}  0.1, & \text{if } x = 0 \\  cx, & \text{if } x = 1 \text{ or } x = 2 \\  c(5 - x), & \text{if } x = 3 \text{ or } x = 4 \\  0, & \text{otherwise} \end{cases}\)
\(\text{Match List-I with List-II:}\)

• A. (A)- (I), (B)- (II), (C)- (III), (D)- (IV)
• B. (A)- (IV), (B)- (III), (C)- (II), (D)- (I)
• C. (A)- (II), (B)- (IV), (C)- (I), (D)- (III)
• D. (A)- (III), (B)- (IV), (C)- (I), (D)- (II)

Hint:

When working with probability distributions, always ensure that the sum of all probabilities equals 1. When solving for unknown probabilities, use the given conditions and simplify the equations step-by-step. It's important to confirm the validity of each probability and use correct substitutions to avoid errors. Additionally, for cumulative probabilities, remember to add up the relevant probabilities as required by the problem.

Answer 1

The Correct Option is B

The total probability must sum to 1:

\[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1. \]

Substitute the provided probabilities:

\[ 0.1 + c(1) + c(2) + c(2) + c(1) = 1. \]

Simplify the equation:

\[ 0.1 + 6c = 1 \Rightarrow 6c = 0.9 \Rightarrow c = 0.15. \]

(A) \( c = 0.15 \). Correspondence: (A) → (IV).

(B) \( P(X \leq 2) \) is the sum of probabilities for \( X=0, X=1, \) and \( X=2 \):

\[ P(X \leq 2) = 0.1 + c(1) + c(2) = 0.1 + 0.15 + 0.3 = 0.55. \]

Correspondence: (B) → (III).

(C) \( P(X = 2) \) is calculated as \( c(2) \):

\[ P(X = 2) = c(2) = 0.3. \) Correspondence: (C) → (II).

(D) \( P(X \geq 2) \) is the sum of probabilities for \( X=2, X=3, \) and \( X=4 \):

\[ P(X \geq 2) = c(2) + c(2) + c(1) = 0.3 + 0.3 + 0.15 = 0.75. \]

Correspondence: (D) → (I).

(A) - (IV), (B) - (III), (C) - (II), (D) - (I).