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

Answer 1

Answer: (D)

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

Answer 2

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)