Question

A random variable X has the following probability distribution:

X	1	2	3	4	5	6	7
P(X)	k	2k	2k	3k	k2	2k2	7k2 + k

Match the options of List-I to List-II:

List-I	List-II
(A) k	(I) 7/10
(B) P(X < 3)	(II) 53/100
(C) P(X ≥ 2)	(III) 1/10
(D) P(2 < X ≤ 7)	(IV) 3/10

Choose the correct answer from the options given below.

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

Answer 1

The Correct Option is D

We know that the sum of all probabilities \(P(X)\) must equal 1:

\(k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1.\)

Simplify the equation:

\(8k + 10k^2 = 1.\)

Dividing through by 2:

\(4k + 5k^2 = \frac{1}{2}.\)

Solve the quadratic equation:

\(5k^2 + 4k - \frac{1}{2} = 0.\)

Using the quadratic formula:

\(k = \frac{-4 \pm \sqrt{4^2 - 4(5)(-\frac{1}{2})}}{2(5)} = \frac{-4 \pm \sqrt{16 + 10}}{10} = \frac{-4 \pm \sqrt{26}}{10}.\)

Since \(k>0\), we take:

\(k = \frac{-4 + \sqrt{26}}{10}.\)

Substituting \(k\), compute probabilities:

\(P(X<3) = P(1) + P(2) = k + 2k = 3k.\)

\(P(X>2) = P(3) + P(4) + P(5) + P(6) + P(7) = 2k + 3k + k^2 + 2k^2 + (7k^2 + k).\)

\(P(2<X<7) = P(3) + P(4) + P(5) + P(6).\)

Matching each value to the given List-II:

\(k = \frac{1}{10}\) (Option III)

\(P(X<3) = \frac{3}{10}\) (Option IV)

\(P(X>2) = \frac{7}{10}\) (Option I)

\(P(2<X<7) = \frac{53}{100}\) (Option II)

Final Matching: (A) - (III), (B) - (IV), (C) - (I), (D) - (II)