Answer:
a) The value of A = 2
b) The value of [tex]B = \dfrac{1}{12}[/tex]
Step-by-step explanation:
a)
Given that:
X should be the random variable that assumes only positive integer values.
The probability function; [tex]P[X = n] = \dfrac{A}{3^n}[/tex] for some constant A and n ≥ 1.
Then, let [tex]\sum \limits ^{\infty}_{n =1} P[X =n] = 1[/tex]
This implies that:
[tex]A \sum \limits ^{\infty}_{n =1} \dfrac{1}{3^n}= 1[/tex]
[tex]A \times \dfrac{\dfrac{1}{3}}{1 - \dfrac{1}{3}} = 1[/tex]
[tex]A \times \dfrac{\dfrac{1}{3}}{\dfrac{2}{3}} = 1[/tex]
[tex]A \times \dfrac{1}{2}=1[/tex]
A = 2
Thus, the value of A = 2
b)
Suppose X represents a e constant A (n> 1). Find A.
b) Let X be a continuous random variable that can assume values between 0 and 3
Then, the density function of x is:
[tex]f_x(x) = \left \{ {{B(x^2+1)} \ \ \ 0 \le x \le 3 \ \ \ \atop {0} \ \ \ otherwise} \right.[/tex]
where; B is constant.
Then, using the property of the probability density function:
[tex]\int ^3_0 \ B (x^2+1 ) \ dx = 1\\[/tex]
Taking the integral, we have:
[tex]B \Big [\dfrac{x^3}{3} +x \Big ]^3_0 = 1[/tex]
[tex]B \Big [\dfrac{3^3}{3} +3 \Big ]= 1[/tex]
[tex]B \Big [\dfrac{27}{3} +3 \Big ] = 1[/tex]
B [ 9 +3 ] = 1
B [ 12 ] = 1
Divide both sides by 12
[tex]B = \dfrac{1}{12}[/tex]
