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.
\(P(X=x)=\begin{cases}0.1, & \text{if } x=0\\ cx, & \text{if } x=1 \text{ or } x=2\\ c(5-x), & \text{if } x=3 \text{ or } x=4\\ 0, & \text{otherwise}\end{cases}\)
Match List-I with List-II:
| List-I | List-II | ||
| (A) | c | (I) | 0.75 |
| (B) | \(P(X \le 2)\) | (II) | 0.3 |
| (C) | \(P(X = 2)\) | (III) | 0.55 |
| (D) | \(P(X \ge 2)\) | (IV) | 0.15 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
The probabilities of the values 0, 1, 2, 3, 4 are:
\(P(0) = 0.1,\ P(1) = c,\ P(2) = 2c\)
\(P(3) = c(5-3) = 2c,\ P(4) = c(5-4) = c\)
(A) c
This matches (IV) 0.15. In a probability distribution the probabilities add up to 1:
\(0.1 + c + 2c + 2c + c = 1\)
\(6c = 0.9\)
\(c = 0.15\)
(B) \(P(X \le 2)\)
This matches (III) 0.55.
\(P(X \le 2) = 0.1 + c + 2c = 0.1 + 0.45 = 0.55\)
(C) \(P(X = 2)\)
This matches (II) 0.3.
\(P(X = 2) = 2c = 0.3\)
(D) \(P(X \ge 2)\)
This matches (I) 0.75.
\(P(X \ge 2) = 2c + 2c + c = 5c = 0.75\)
So the matching is (A) - (IV), (B) - (III), (C) - (II), (D) - (I).