Respuesta :
A) Since there are a total of four qualifiers and we need to know the probability that 1 person wins both prizes, we will first find the probability of one person winning the first draw. That would be:
[tex]\begin{gathered} P=\frac{\text{ number of winners}}{\text{ total number of qualifiers}} \\ P_1=\frac{1}{4} \end{gathered}[/tex]Now, since one person can be drawn again, the same would apply to the second prize:
[tex]\begin{gathered} P=\frac{\text{numberofw\imaginaryI nners}}{\text{totalnumberofqual\imaginaryI f\imaginaryI ers}} \\ P_2=\frac{1}{4} \end{gathered}[/tex]Therefore, the probability that one person wins both would be:
[tex]\begin{gathered} P=\frac{1}{4}\times\frac{1}{4} \\ P=\frac{1}{16} \end{gathered}[/tex]B) Next, we will find the probability that there will be 2 different winners. First, we find the probability of 1 person winning the prize:
[tex]P=\frac{1}{4}[/tex]Then, the probability of another person winning the prize:
[tex]P=\frac{1}{4}[/tex]The probability of having 2 different winners would then be:
[tex]\begin{gathered} P=(\frac{1}{4}\times\frac{1}{4})+(\frac{1}{4}\times\frac{1}{4}) \\ P=\frac{1}{8} \end{gathered}[/tex]C) Now, the probability of Sofia winning at least one prize. The probability of winning at least 1 prize is equal to 1 - the probability of winning no prizes at all:
[tex]\begin{gathered} P=1-(\frac{3}{4})^2 \\ P=\frac{7}{16} \end{gathered}[/tex]D) The answer for D would be the same for question A. Given that Frank is the one person who wins both prizes:
[tex]\begin{gathered} P=\frac{1}{4}\times\frac{1}{4} \\ P=\frac{1}{16} \end{gathered}[/tex]E) For E, this would be the same situation as B, given that the two different winners are Jake and Eldridge:
[tex]\begin{gathered} P=(\frac{1}{4}\times\frac{1}{4})+(\frac{1}{4}\times\frac{1}{4}) \\ P=\frac{1}{8} \end{gathered}[/tex]