Question

The random variable \(\boldsymbol{n}\) represents the number of units of a product sold per day in a store. The probability distribution of \(n\) is given by \(P(n) .\) Find the probability that two units are sold in a given day \([P(2)]\) and show that \(P(1)+P(2)+P(3)+\cdots=1\). $$ P(n)=\frac{1}{3}\left(\frac{2}{3}\right)^{n} $$

Short answer

The probability of selling two units in a day is \( \frac{4}{27} \), and it is shown that the sum of probabilities of all outcomes in the distribution equals 1.

Step-by-step solution

A. Calculate the Probability of Selling 2 Units in a Day

To find P(2), simply substitute n=2 into the given formula. \nSyntax: \[ P(2)=\frac{1}{3}\left(\frac{2}{3}\right)^{2} \]

B. Evaluate P(2)

Perform the arithmetic to compute P(2). \[ P(2)=\frac{1}{3}\times\left(\frac{4}{9}\right)=\frac{4}{27} \]

C. Prove that Sum of All Probabilities Equals 1

Here, this can be represented by the formula for the sum of an infinite geometric series, \( a/(1-r) \), where a is the first term and r the common ratio, which in this case a=1/3 and r=2/3. Substituting these values into the formula gives: \[ \frac{1/3}{1-(2/3)}=1 \]