Question

A baker man sells 5 types of cakes. Profits due to the sale of each type of cake are respectively Rs. 3, Rs. 2.5, Rs. 2, Rs. 1.5, and Rs. 1. The demands for these cakes are 10%, 5%, 20%, 50% and 15% respectively. What is the expected profit per cake? \[\]

Answer 1

Hint: We take a discrete random variable $X$ and assign the profits as outcomes $X$ takes as $X={{x}_{1}}=3,X={{x}_{2}}=2.5,X={{x}_{3}}=2,X={{x}_{4}}=1.5,X={{x}_{5}}=1$ and take the demands as the corresponding probabilities ${{p}_{1}},{{p}_{2}},{{p}_{3}},{{p}_{4}},{{p}_{5}}$. We calculate the expected profit using the formula $E\left[ X \right]=\sum\limits_{k=1}^{n}{{{p}_{k}}{{x}_{k}}}$ for $n=5.$\[\]

Complete step by step answer:
We know that the expected value of a random variable $X$ is the weighted average of all possible outcomes that $X$can take. If there are $k$outcomes for the random variable $X$ denoted as $X={{x}_{1}},X={{x}_{2}},...,X={{x}_{k}}$ with corresponding probabilities $P\left( X={{x}_{1}} \right)={{p}_{1}},P\left( X={{x}_{2}} \right)={{p}_{2}},...,P\left( X={{x}_{k}} \right)={{p}_{k}}$ then the expectation of $X$ is denoted as $E\left[ X \right]$ and given as
\[E\left[ X \right]={{p}_{1}}{{x}_{1}}+{{p}_{2}}{{x}_{2}}+...+{{p}_{k}}{{x}_{k}}\]
The probabilities are conditioned by their sum as ${{p}_{1}}+{{p}_{2}}+...+{{p}_{k}}=1$. The probabilities are also called weights. The weighted average or the expectation measures how much value is assigned per outcome. \[\]

We are given in the question that the baker man sells 5 types of cakes, so there are total 5 outcomes which we denote as ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}}$. Profits due to the sale of each type of cake are respectively Rs. 3, Rs. 2.5, Rs. 2, Rs. 1.5, and Rs. 1. Let us define a random variable $X$ which takes the profit as outcomes. So we have the outcomes
\[X={{x}_{1}}=3,X={{x}_{2}}=2.5,X={{x}_{3}}=2,X={{x}_{4}}=1.5,X={{x}_{5}}=1\]
We are further given in the question that the demands for these cakes are 10%, 5%, 20%, 50% and 15% respectively. The higher the demand, the higher is the probability of selling the cakes. So we can take the demand as probabilities. So we have
\[\begin{gathered}
  P\left( {X = 3} \right) = {p_1} = 10\% = \dfrac{{10}}{{100}} = 0.1 \\
  P\left( {X = 2.5} \right) = {p_2} = 5\% = \dfrac{5}{{100}} = .05 \\
  P\left( {X = 2} \right) = {p_3} = 20\% = \dfrac{{20}}{{100}} = 0.2 \\
  P\left( {X = 1.5} \right) = {p_4} = 50\% = \dfrac{{50}}{{100}} = .5 \\
  P\left( {X = 1} \right) = {p_5} = 15\% = \dfrac{{15}}{{100}} = .15 \\
\end{gathered} \]
Let us check whether they satisfy condition on sum of probabilities
\[{{p}_{1}}+{{p}_{2}}+{{p}_{3}}+{{p}_{4}}+{{p}_{5}}=0.1+0.05+.2+.5+.15=1\]
 So the probabilities can be assigned in the discrete distribution for $X.$ Now we find the expectation of selling of cakes as
\[\begin{align}
  & E\left[ X \right]={{p}_{1}}{{x}_{1}}+{{p}_{2}}{{x}_{2}}+{{p}_{3}}{{x}_{3}}+{{p}_{4}}{{x}_{4}}+{{p}_{5}}{{x}_{5}} \\
 & \Rightarrow E\left[ X \right]=3\times 0.1+2.5\times 0.05+2\times 0.2+1.5\times 0.5+1\times .15 \\
 & \Rightarrow E\left[ X \right]=0.3+0.125+0.4+.75+0.15 \\
 & \Rightarrow E\left[ X \right]=1.725 \\
\end{align}\]

So the expected profit per cake is 1.725 rupees.

Note: If all probabilities are equal that is ${{p}_{1}}={{p}_{2}}=...={{p}_{k}}$ then we call the outcomes ${{x}_{1}},{{x}_{2}},...,{{x}_{k}}$ equiprobable and the expectation becomes simple average also known as arithmetic mean. If $X$ is a continuous random variable the expectation is given by $E\left[ X \right]=\int\limits_{R}{xf\left( x \right)}$ where $f\left( x \right)$ is the probability distribution function.