Given:
• Top cash prize = $697
,
• First set of numbers to pick 4 different numbers from = 1 to 53
,
• Second set to pick one number from = 1 to 46
,
• Minimum award = $225
Given that the player wins the minimum award by matching three numbers drawn from 1 to 53 and matching number through 1 to 43, let's find the probability of winning the minimum award.
To find the probability, we have the following:
Number of ways to pick 4 different numbers from 1 to 53:
[tex]\begin{gathered} ^{53}C_4=\frac{53!}{4!(53-4)!}=\frac{53!}{4!*49!} \\ \\ ^{53}C_4=\frac{53*52*51*50*49!}{4!*49!}=\frac{53*52*51*50}{4*3*2*1}=\frac{7027800}{24}=292825 \end{gathered}[/tex]
Now, the probability of matching 3 numbers from the 4 different numbers picked and matching the number on the gold ball (1 to 46) will be:
[tex]\begin{gathered} P=\frac{^4C_3*(53-4)}{^{53}C_4}*\frac{1}{46} \\ \\ P=\frac{\frac{4!}{3!(4-3)!}*49}{292825}*\frac{1}{46} \\ \\ P=\frac{\frac{4*3!}{3!*1!}*49}{292825}*\frac{1}{46} \\ \\ P=\frac{4*49}{292825}*\frac{1}{46} \\ \\ P=\frac{196*1}{292825*46} \\ \\ P=\frac{196}{13469950} \\ \\ P=\frac{98}{6734975} \end{gathered}[/tex]
Therefore, the probability of winning a minimum award is:
[tex]\frac{98}{6734975}[/tex]
• ANSWER:
[tex]\frac{98}{6734975}[/tex]
