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)

Hint:

When solving probability problems like this, make sure to work step by step. Break down the quadratic equation using the quadratic formula and check that your substitutions are accurate when calculating probabilities. Also, it’s useful to keep track of the values you substitute and the intermediate steps to ensure that everything adds up correctly.

Answer 1

The Correct Option is D

The sum of all probabilities \(P(X)\) must equal 1:

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

The simplified equation is:

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

Dividing by 2 yields:

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

Rearranging into a quadratic equation:

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

Applying the quadratic formula:

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

Given that \(k>0\), the value of \(k\) is:

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

Substituting the value of \(k\), the probabilities are computed as follows:

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

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

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

Matching the computed values to the provided List-II:

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

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

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

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

The final matching is: (A) - (III), (B) - (IV), (C) - (I), (D) - (II).