Q.4.33 - A First Course in Probability

Question

A newsboy purchases papers at $10$ cents and sells them at $15$ cents. However, he is not allowed to return unsold papers. If his daily demand is a binomial random variable with $n=10,p=\frac{1}{3}$, approximately how many papers should he purchase so as to maximize his expected profit?

Step-by-Step Solution

Answer

The newspaper boy should need $3$ more to maximize the profit.

Step 1: Given information

A newsboy purchases papers at $10$ cents and sells them at $15$ cents. However, he is not allowed to return unsold papers.

Step 2: Solution

Let,

$m=$Paper purchased by newsboy

$n=10$

$m⩽10$

$p=$profit

$D=$demand of the day

So,

$E\left(P\right)=\sum _{k=0}^{10} 15\mathrm{Min}\{k,m\}P(D=k)-10m$

$=15\sum _{k=0}^{m-1} kP(D=k)+15m\sum _{k=m}^{10} P(D=k)-10m$

Let $f\left(m\right)$ be right hand side of above equation.

Therefore,

$f(m+1)-f\left(m\right)=\left\{\begin{array}{l}15\sum _{k=0}^m kP(D=k)-15\sum _{k=0}^{m-1} kP(D=k)\\ +15(m+1)\sum _{k=m+1}^{10} P(D=k)-15m\sum _{k=m}^{10} P(D=k)-10\end{array}\right\}$

$=15\sum _{k=m+1}^{10} P(D=k)-10$

$=15\left(1-\sum _{k=0}^m P(D=k)\right)-10$

$=5-15\sum _{k=0}^m P(D=k)$

Step 3: Final solution

Now we need to find the maximum value $m$

$F_D\left(m\right)=\sum _{k=0}^m P(D=k)$

$=\sum _{k=0}^m \left(\begin{array}{l}10\\ k\end{array}\right){\left(\frac{1}{3}\right)}^k{\left(\frac{2}{3}\right)}^{10-k}\le \frac{1}{3}$

If $m=2$

$F_D\left(2\right)=\sum _{k=0}^2 \left(\begin{array}{l}10\\ k\end{array}\right){\left(\frac{1}{3}\right)}^k{\left(\frac{2}{3}\right)}^{10-k}$

$=\left(\begin{array}{l}10\\ 0\end{array}\right){\left(\frac{1}{3}\right)}^0{\left(\frac{2}{3}\right)}^{10-0}+\left(\begin{array}{l}10\\ 1\end{array}\right){\left(\frac{1}{3}\right)}^1{\left(\frac{2}{3}\right)}^{10-1}+\left(\begin{array}{l}10\\ 2\end{array}\right){\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^{10-2}$

$=0.01735+0.08674+0.19513$

$=0.2992\left(<\frac{1}{3}\right)$

If $m=3$

$F_D\left(3\right)=\sum _{k=0}^3 \left(\begin{array}{l}10\\ k\end{array}\right){\left(\frac{1}{3}\right)}^k{\left(\frac{2}{3}\right)}^{10-k}$

$=\left\{\begin{array}{l}\left(\begin{array}{c}10\\ 0\end{array}\right){\left(\frac{1}{3}\right)}^0{\left(\frac{2}{3}\right)}^{10-0}+\left(\begin{array}{c}10\\ 1\end{array}\right){\left(\frac{1}{3}\right)}^1{\left(\frac{2}{3}\right)}^{10-1}+\left(\begin{array}{c}10\\ 2\end{array}\right){\left(\frac{1}{3}\right)}^2{\left(\frac{2}{3}\right)}^{10-2}\\ +\left(\begin{array}{c}10\\ 3\end{array}\right){\left(\frac{1}{3}\right)}^3{\left(\frac{2}{3}\right)}^{10-3}\end{array}\right\}$

$=0.01735+0.08674+0.19513+0.26014$

$=0.5594\left(>\frac{1}{3}\right)$

Since $F_D$ is an increasing function we concluded that $F_D\left(m\right)<\frac{1}{3}$ for all $m\le 2$ and $F_D\left(m\right)>\frac{1}{3}$ for all $m\ge 3$.

If, $f(m+1)-f\left(m\right)>0$ for all $m=0,1,2$ and $f(m+1)-f\left(m\right)<0$ for all $m=3,\dots .,9$

This means $f$ gets its maximum when $m=3$

So, the newspaper boy should need $3$ more to maximize the profit.

Step 4: Final answer

The newspaper boy should need $3$ more to maximize the profit.