To determine the probability that a randomly selected purchase includes at least two tickets, we need to consider the probabilities associated with buying 2, 3, 4, or 5 tickets.
From the provided table:
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
Number of \\
Tickets
\end{tabular} & 1 & 2 & 3 & 4 & 5 \\
\hline Probability & 0.17 & 0.55 & 0.20 & 0.06 & 0.02 \\
\hline
\end{tabular}
\][/tex]
The probabilities are:
- Probability of buying 1 ticket: [tex]\(0.17\)[/tex]
- Probability of buying 2 tickets: [tex]\(0.55\)[/tex]
- Probability of buying 3 tickets: [tex]\(0.20\)[/tex]
- Probability of buying 4 tickets: [tex]\(0.06\)[/tex]
- Probability of buying 5 tickets: [tex]\(0.02\)[/tex]
The issue at hand asks us to find the probability of purchasing at least 2 tickets. This requires us to sum the probabilities associated with purchasing 2, 3, 4, or 5 tickets:
[tex]\[
P(\text{at least 2 tickets}) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\][/tex]
We now add the probabilities together:
[tex]\[
P(\text{at least 2 tickets}) = 0.55 + 0.20 + 0.06 + 0.02
\][/tex]
Summing these probabilities:
[tex]\[
0.55 + 0.20 = 0.75
\][/tex]
[tex]\[
0.75 + 0.06 = 0.81
\][/tex]
[tex]\[
0.81 + 0.02 = 0.83
\][/tex]
Thus, the probability that a randomly selected purchase has at least two tickets is [tex]\(0.83\)[/tex].
Therefore, the answer is [tex]\(0.83\)[/tex].
