Question

If probability distribution is given by \[ P(x) = \begin{array}{c|c|c|c|c|c|c|c} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & k & 2k^2 & 6k^2 & 2k^2 + k & 4k & k & k \\ \end{array} \] Then, the value of \( P(3<x \leq 6) \) is:

• A. 0.6
• B. 0.8
• C. 0.4
• D. 0.2

Hint:

When dealing with probability distributions, always ensure that the total probability sums to 1, and use this to solve for unknown constants.

Answer 1

The Correct Option is A

The problem asks us to find the probability \( P(3<x \leq 6) \) from the given probability distribution. Let's solve it step by step:

x	0	1	2	3	4	5	6	7
P(x)	\(k\)	\(2k^2\)	\(6k^2\)	\(2k^2 + k\)	\(4k\)	\(k\)	\(k\)

First, we need to find the value of \(k\). The sum of probabilities of all outcomes must equal 1. Thus, we have:

\(k + 2k^2 + 6k^2 + (2k^2 + k) + 4k + k + k = 1\)

Combine like terms:

\(16k^2 + 7k = 1\)

This is a quadratic equation in terms of \(k\):

\(16k^2 + 7k - 1 = 0\)

Solving this quadratic equation using the quadratic formula:

\(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 16\), \(b = 7\), \(c = -1\). Calculate the discriminant:

\(b^2 - 4ac = 7^2 - 4(16)(-1) = 49 + 64 = 113\)

Therefore:

\(k = \frac{-7 \pm \sqrt{113}}{32}\)

To find the valid probability, we take the positive value:

Approximating the square root, we have:

\(k \approx \frac{-7 + 10.630}{32}\)

\(k \approx \frac{3.63}{32} \approx 0.1134\)

Now we calculate \( P(3<x \leq 6) \):

For \( 3 < x \leq 6 \), the values of \(x\) are \(4, 5, 6\).

Thus:

\(P(4) = 4k, \, P(5) = k, \, P(6) = k\)

So, \( P(3 < x \leq 6) = 4k + k + k = 6k \).

\(P(3 < x \leq 6) = 6 \times 0.1134 = 0.6804 \approx 0.6\)

Therefore, the value of \( P(3

This matches with option 0.6, which is the correct answer.