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

\begin{tabular}{lccccc}
Payout (\[tex]$) & 3 & 4 & 5 & 6 & 7 \\
Probability & 0.01 & 0.04 & 0.10 & 0.20 & 0.65
\end{tabular}

Expected Value $[/tex]=[?]$



Answer :

To find the expected value of the winnings from the game, you need to follow the steps for calculating the expected value of a discrete random variable. The expected value (E) is calculated by multiplying each possible payout by its corresponding probability and then summing up all these products.

Let's denote the payouts as [tex]\( X_i \)[/tex] and the probabilities as [tex]\( P(X_i) \)[/tex].
Here are the payouts and their corresponding probabilities:

[tex]\[ \begin{array}{c|c|c|c|c|c} \text{Payout (\$)} & 3 & 4 & 5 & 6 & 7 \\ \hline \text{Probability} & 0.01 & 0.04 & 0.10 & 0.20 & 0.65 \\ \end{array} \][/tex]

The expected value [tex]\( E(X) \)[/tex] is computed as:
[tex]\[ E(X) = \sum_{i} X_i \cdot P(X_i) \][/tex]

Calculating each term individually:

1. For payout [tex]\( X_1 = 3 \)[/tex]:
[tex]\[ 3 \times 0.01 = 0.03 \][/tex]

2. For payout [tex]\( X_2 = 4 \)[/tex]:
[tex]\[ 4 \times 0.04 = 0.16 \][/tex]

3. For payout [tex]\( X_3 = 5 \)[/tex]:
[tex]\[ 5 \times 0.10 = 0.50 \][/tex]

4. For payout [tex]\( X_4 = 6 \)[/tex]:
[tex]\[ 6 \times 0.20 = 1.20 \][/tex]

5. For payout [tex]\( X_5 = 7 \)[/tex]:
[tex]\[ 7 \times 0.65 = 4.55 \][/tex]

Now, sum these values together to find the expected value:
[tex]\[ E(X) = 0.03 + 0.16 + 0.50 + 1.20 + 4.55 = 6.44 \][/tex]

Therefore, the expected value of the winnings from the game is:
[tex]\[ \boxed{6.44} \][/tex]