Since 1 ticket has a payoff of $ 630 the probability of winning $ 630 is:
[tex]\frac{1}{100}=0.01[/tex]then,
[tex]P(630)=0.01[/tex]Since the 99 remaining tickets have a payoff of $0, the probability of winning $0 is:
[tex]P(0)=0.99[/tex]To find E(x), the expected value of x, find k⋅P(k) for each value k that x can take. Then sum these terms.
[tex]\begin{gathered} x\times P(x)=0\times0.00=0 \\ x\times P(x)=630\times0.01=6.3 \end{gathered}[/tex]therefore:
[tex]E(x)=0+6.3=6.3[/tex]this is the expected payoff is $6.3
answer: $6.3