Question

George will play a game at the school fair in which he will toss a penny, a dime, and a quarter at the same time. He will receive $3$ points for each coin that lands with heads face up. Let a represent the total number of points awarded on any toss of the coins. What is the expected value of a?

Answer 1

Hint: Since three coins are tossed, the sample space will be given as $\left\{ HHH,HHT,HTH,THH,THT,TTH,HTT,TTT \right\}$. Since the total outcomes are equal to eight, each of which is equally likely to occur, each outcome will have a probability of $\dfrac{1}{8}$. According to the question, the variable a denotes the points awarded to George, and $3$ points are awarded for each head. Therefore, the possible values of a according to the number of heads in the sample space will become $\left\{ 9,6,6,6,3,3,3,0 \right\}$. Finally, the mean of a can be calculated using the formula $E\left( X \right)=\sum\limits_{i=1}^{n}{{{X}_{i}}{{P}_{i}}}$.

Complete step-by-step answer:
Since three coins are tossed simultaneously, the total number of possible outcomes will be equal to ${{2}^{3}}$ which is equal to $8$. The sample space for this experiment can be given by the set
$\Rightarrow S=\left\{ HHH,HHT,HTH,THH,THT,TTH,HTT,TTT \right\}$
Now, according to the above question, George receives three points for each head obtained. Thus, according to the number of heads in each of the outcomes, we can assign points to each of the outcomes in the above sample space and write the corresponding values of a as
$\Rightarrow a=\left\{ 9,6,6,6,3,3,3,0 \right\}$
Now, we know that the mean or the expectation of a random variable X is given by
$\Rightarrow E\left( X \right)={{X}_{1}}{{P}_{1}}+{{X}_{2}}{{P}_{2}}+......$
Therefore, the mean of the variable a will be given by
$\Rightarrow E\left( a \right)={{a}_{1}}{{P}_{1}}+{{a}_{2}}{{P}_{2}}+......$
Putting the values of a from (i) we get
\[\Rightarrow E\left( a \right)=9{{P}_{1}}+6{{P}_{2}}+6{{P}_{3}}+6{{P}_{4}}+3{{P}_{5}}+3{{P}_{6}}+3{{P}_{7}}+\left( 0 \right){{P}_{8}}\]
Since each of the outcomes is equally likely to occur and has the probability of $\dfrac{1}{8}$, we can substitute \[{{P}_{1}}={{P}_{2}}=.......=\dfrac{1}{8}\] in the above equation to get
\[\begin{align}
  & \Rightarrow E\left( a \right)=\dfrac{9}{8}+\dfrac{6}{8}+\dfrac{6}{8}+\dfrac{6}{8}+\dfrac{3}{8}+\dfrac{3}{8}+\dfrac{3}{8}+\dfrac{0}{8} \\
 & \Rightarrow E\left( a \right)=\dfrac{36}{8} \\
 & \Rightarrow E\left( a \right)=\dfrac{9}{2} \\
\end{align}\]
Hence, the expected value of a is equal to \[\dfrac{9}{2}\].

Note: We must not be confused why the expected value of a is equal to a fractional number, when the points scored by George are in the multiples of three. Since it is the average value and not the absolute value, it can be a fraction.