A random variable takes values \(0,1,2,3 \ldots \ldots\) with probability proportional to \((x+1)(1 / 5)^{x}\). Then (a) \(P(x=0)=(16 / 25)\) (b) \(P(x \geq 1)=(16 / 25)\) (c) \(P(x \geq 1)=(7 / 25)\) (d) none

Since the probabilities must add up to 1, we can write: \(1 = c\sum_{x=0}^\infty (x+1)\left(\frac{1}{5}\right)^x\) Let's calculate this sum: \(\sum_{x=0}^\infty (x+1)\left(\frac{1}{5}\right)^x = 0\cdot\left(\frac{1}{5}\right)^0+1\cdot\left(\frac{1}{5}\right)^1+2\cdot\left(\frac{1}{5}\right)^2+3\cdot\left(\frac{1}{5}\right)^3+\ldots\)

This sum is a modified geometric series, which can be written as: \(\left(\frac{1}{5}\right)^0\left(0\cdot1\right)+\left(\frac{1}{5}\right)^1\left(1\cdot1\right)+\left(\frac{1}{5}\right)^2\left(2\cdot1\right)+\left(\frac{1}{5}\right)^3\left(3\cdot1\right)+\ldots\) Notice that the series of coefficients is not constant (it increases by 1 at each term). However, we can rewrite the series as: \(\sum_{x=0}^\infty x\left(\frac{1}{5}\right)^x + \sum_{x=0}^\infty 1\left(\frac{1}{5}\right)^x\) Now, the second sum is a geometric series, with common ratio \(r=\frac{1}{5}\). We can find its sum using the following formula: \(S = \frac{a}{1-r}\) where \(S\) is the sum of the series, \(a\) is the first term in the series and \(r\) is the common ratio. So for the second sum: \(S = \frac{1}{1-\frac{1}{5}} = \frac{1}{\frac{4}{5}} = \frac{5}{4}\) For the first sum, after multiplying each term by the common ratio, differentiating with respect to the common ratio, and evaluating at the given common ratio, we will obtain the sum as \(S = \frac{1}{(1-r)^2}\): \(S = \frac{1}{\left(1-\frac{1}{5}\right)^2} = \frac{1}{\frac{16}{25}} = \frac{25}{16}\) Now, let's add the two sums: \(\sum_{x=0}^\infty (x+1)\left(\frac{1}{5}\right)^x = \frac{25}{16} + \frac{5}{4} = \frac{33}{16}\)

Now we can find the constant of proportionality, c: \(1 = c \cdot \frac{33}{16}\) To solve for c, we divide both sides by \(\frac{33}{16}\): \(c = \frac{16}{33}\)

Now we can calculate the probabilities using the constant of proportionality: (a) \(P(x=0) = \frac{16}{33} \cdot (0+1)\left(\frac{1}{5}\right)^0 = \frac{16}{33}\) (b) We can calculate the probability \(P(x\geq 1)\) by using the formula for the sum of an infinite geometric series, but excluding the term for x = 0: \(P(x \geq 1) = 1 - P(x=0) = 1 - \frac{16}{33} = \frac{17}{33}\) Now we compare our calculated probabilities with the given statements: (a) \(P(x=0) = \frac{16}{33} \neq\frac{16}{25}\) (b) \(P(x \geq 1) = \frac{17}{33} \neq\frac{16}{25}\) (c) \(P(x \geq 1) = \frac{17}{33} \neq\frac{7}{25}\) Since none of the given statements match our calculated probabilities, the correct answer is: (d) none