Find the expected value of the winnings from a game that has the following payout probability distribution:

\begin{tabular}{c|ccccc}
Payout [tex]$(\$[/tex])[tex]$ & 1 & 2 & 3 & 4 & 5 \\
\hline
Probability & 0.19 & 0.13 & 0.37 & 0.10 & 0.21
\end{tabular}

Expected Value $[/tex]=[tex]$ $[/tex]\square$

Round to the nearest hundredth.



Answer :

To compute the expected value (EV) of a game with a given payout probability distribution, we use the formula for the expected value of a discrete random variable:
[tex]\[ EV = \sum (payout \times probability) \][/tex]

Given the payout and probability distribution in the table:
[tex]\[ \begin{tabular}{c|ccccc} Payout $(\$)$ & 1 & 2 & 3 & 4 & 5 \\ \hline Probability & 0.19 & 0.13 & 0.37 & 0.10 & 0.21 \end{tabular} \][/tex]

Let's calculate the EV step-by-step:

1. Multiply each payout by its corresponding probability:
- For a payout of \[tex]$1: \(1 \times 0.19 = 0.19\) - For a payout of \$[/tex]2: [tex]\(2 \times 0.13 = 0.26\)[/tex]
- For a payout of \[tex]$3: \(3 \times 0.37 = 1.11\) - For a payout of \$[/tex]4: [tex]\(4 \times 0.10 = 0.40\)[/tex]
- For a payout of \$5: [tex]\(5 \times 0.21 = 1.05\)[/tex]

2. Add all these products together to get the expected value:
[tex]\[ EV = 0.19 + 0.26 + 1.11 + 0.40 + 1.05 = 3.01 \][/tex]

The expected value of the winnings from this game is:
[tex]\[ 3.01 \][/tex]

Rounding the result to the nearest hundredth doesn't change its value since it is already rounded to two decimal places.

Therefore, the expected value (rounded to the nearest hundredth) is:
[tex]\[ \boxed{3.01} \][/tex]