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Addison sells 100 tickets for \[tex]$10 each for a raffle. There is 1 award for \$[/tex]100, 4 awards for \[tex]$50, and 10 awards for \$[/tex]30. The remaining proceeds go to hosting the contest. Which table correctly displays the probability distribution?

[tex]\[
\begin{tabular}{|c|c|}
\hline \multicolumn{2}{|c|}{Distribution of Awards} \\
\hline Prize & $P(x)$ \\
\hline none & 0.85 \\
\hline \$30 & 0.10 \\
\hline \$50 & 0.04 \\
\hline \$100 & 0.01 \\
\hline
\end{tabular}
\][/tex]


Sagot :

To determine if the given table displays the correct probability distribution for the raffle prizes, we need to calculate the probabilities for each prize category and compare them with the given table.

We know there are 100 tickets sold, and the prizes are distributed as follows:
- 1 award of \[tex]$100 - 4 awards of \$[/tex]50
- 10 awards of \[tex]$30 - The remaining tickets do not win any award. First, let's calculate the probability of winning each prize: 1. Probability of winning \$[/tex]100:
There is 1 ticket that wins \[tex]$100 out of 100 tickets. \[ P(\$[/tex]100) = \frac{1}{100} = 0.01 \]

2. Probability of winning \[tex]$50: There are 4 tickets that win \$[/tex]50 out of 100 tickets.
[tex]\[ P(\$50) = \frac{4}{100} = 0.04 \][/tex]

3. Probability of winning \[tex]$30: There are 10 tickets that win \$[/tex]30 out of 100 tickets.
[tex]\[ P(\$30) = \frac{10}{100} = 0.10 \][/tex]

4. Probability of winning nothing:
The remaining tickets do not win any prize. The number of tickets that win nothing can be calculated as follows:
[tex]\[ \text{Tickets winning nothing} = 100 - (1 + 4 + 10) = 100 - 15 = 85 \][/tex]
Thus,
[tex]\[ P(\text{none}) = \frac{85}{100} = 0.85 \][/tex]

Now, let's compare these computed probabilities with the table:

\begin{tabular}{|c|c|}
\hline \multicolumn{2}{|c|}{\begin{tabular}{c}
Distribution \\
of Awards
\end{tabular}} \\
\hline Prize & [tex]$P(x)$[/tex] \\
\hline none & 0.85 \\
\hline \[tex]$30 & 0.01 \\ \hline \$[/tex]50 & 0.04 \\
\hline \[tex]$100 & 0.10 \\ \hline \end{tabular} ### Comparison - Probability of winning \$[/tex]100:
- Calculated: 0.01
- Given: 0.10
- Probability of winning \[tex]$50: - Calculated: 0.04 - Given: 0.04 - Probability of winning \$[/tex]30:
- Calculated: 0.10
- Given: 0.01
- Probability of winning nothing:
- Calculated: 0.85
- Given: 0.85

Clearly, the probabilities provided in the table for \[tex]$100 and \$[/tex]30 are swapped. The correct probability distribution should be:

\begin{tabular}{|c|c|}
\hline \multicolumn{2}{|c|}{\begin{tabular}{c}
Distribution \\
of Awards
\end{tabular}} \\
\hline Prize & [tex]$P(x)$[/tex] \\
\hline none & 0.85 \\
\hline \[tex]$30 & 0.10 \\ \hline \$[/tex]50 & 0.04 \\
\hline \$100 & 0.01 \\
\hline
\end{tabular}

Therefore, the given table does not correctly display the probability distribution.