Question

Frank, Sofia, Eldridge, and Jake are the four qualifiers for a charity raffle with two $500 prizes. One of their names will be drawn for the first prize then replaced, at which point the second prize winner will be drawn. find the probability thatA) one person wins both prizes Answer =B) There are two different winners Answer =C) Sofia wins at least one prize Answer is =D) Frank wins both prizes Answer is =E) The two winners are Jake and Eldridge Answer is =

Answer 1

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:

`\begin{gathered} P=\frac{\text{ number of winners}}{\text{ total number of qualifiers}} \\ P_1=(1)/(4) \end{gathered}`

Now, since one person can be drawn again, the same would apply to the second prize:

`\begin{gathered} P=\frac{\text{numberofw\imaginaryI nners}}{\text{totalnumberofqual\imaginaryI f\imaginaryI ers}} \\ P_2=(1)/(4) \end{gathered}`

Therefore, the probability that one person wins both would be:

`\begin{gathered} P=(1)/(4)*(1)/(4) \\ P=(1)/(16) \end{gathered}`

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:

`P=(1)/(4)`

Then, the probability of another person winning the prize:

`P=(1)/(4)`

The probability of having 2 different winners would then be:

`\begin{gathered} P=((1)/(4)*(1)/(4))+((1)/(4)*(1)/(4)) \\ P=(1)/(8) \end{gathered}`

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:

`\begin{gathered} P=1-((3)/(4))^2 \\ P=(7)/(16) \end{gathered}`

D) The answer for D would be the same for question A. Given that Frank is the one person who wins both prizes:

`\begin{gathered} P=(1)/(4)*(1)/(4) \\ P=(1)/(16) \end{gathered}`

E) For E, this would be the same situation as B, given that the two different winners are Jake and Eldridge:

`\begin{gathered} P=((1)/(4)*(1)/(4))+((1)/(4)*(1)/(4)) \\ P=(1)/(8) \end{gathered}`