To find the possible value(s) of p for the given probability distribution of the discrete random variable X , we need to ensure that the sum of all probabilities equals 1. The probability distribution is given as follows:

x	1	2	3	4	5
\mathop{\mathrm{P}}\left(X=x\right)	0.1	\frac{\left(2p-3\right)^2}{4}	0.2	0.2	\frac{6p-11}{4}

The sum of all probabilities must be equal to 1 :

0.1 + \frac{(2p-3)^2}{4} + 0.2 + 0.2 + \frac{6p-11}{4} = 1

Simplify the equation:

0.5 + \frac{(2p-3)^2}{4} + \frac{6p-11}{4} = 1

Subtract 0.5 from both sides:

\frac{(2p-3)^2}{4} + \frac{6p-11}{4} = 0.5

Multiply the entire equation by 4 to eliminate the fractions:

(2p-3)^2 + 6p - 11 = 2

Subtract 2 from both sides:

(2p-3)^2 + 6p - 13 = 0

Now, expand (2p-3)^2:

(2p-3)^2 = 4p^2 - 12p + 9

Substitute back into the equation:

4p^2 - 12p + 9 + 6p - 13 = 0

Combine like terms:

4p^2 - 6p - 4 = 0

This is a quadratic equation in the standard form ax^2 + bx + c = 0 , where a = 4 , b = -6 , and c = -4 . We can solve it using the quadratic formula :

p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Calculate the discriminant:

b^2 - 4ac = (-6)^2 - 4 \times 4 \times (-4) = 36 + 64 = 100

Since the discriminant is positive, there are two real solutions:

p = \frac{6 \pm \sqrt{100}}{8}

Calculate the values of p :

p_1 = \frac{6 + 10}{8} = \frac{16}{8} = 2

p_2 = \frac{6 - 10}{8} = \frac{-4}{8} = -0.5

Since probabilities must be non-negative, we need to check which of these values result in valid probabilities for all x .

For p = 2 :
- \frac{(2p-3)^2}{4} = \frac{(2 \times 2 - 3)^2}{4} = \frac{1^2}{4} = 0.25
- \frac{6p-11}{4} = \frac{6 \times 2 - 11}{4} = \frac{1}{4} = 0.25

Both probabilities are non-negative and the sum of all probabilities is:

0.1 + 0.25 + 0.2 + 0.2 + 0.25 = 1

For p = -0.5 :
- \frac{(2p-3)^2}{4} = \frac{(2 \times -0.5 - 3)^2}{4} = \frac{(-4)^2}{4} = 4 (not valid as it exceeds 1)
- \frac{6p-11}{4} = \frac{6 \times -0.5 - 11}{4} = \frac{-14}{4} = -3.5 (not valid as it is negative)

Thus, the possible value of p that satisfies the conditions of the probability distribution is p = 2 .

x	1	2	3	4	5
P(X=x)	0.1	4(2p−3)2	0.2	0.2	46p−11

x	1	2	3	4	5
P(X=x)	0.1	4(2p−3)2	0.2	0.2	46p−11