Respuesta :
Solution
We are given that
[tex]\begin{gathered} p(A)=0.2 \\ p(B)=0.45 \\ p(C)=0.35 \end{gathered}[/tex]Note 1: Probability Formula To use
[tex]p(A\cup B)=p(A)+p(B)-p(A\cap B)[/tex]Note 2: Team A, B and C are Mutually Exclusive
[tex]\begin{gathered} p(A)+p(B)+p(C)=0.2+0.45+0.35=1 \\ Th\text{ey are mutually exclusive} \\ A\cap B=B\cap C=A\cap C=\varnothing \\ p(A\cap B)=p(B\cap C)=p(A\cap C)=0 \end{gathered}[/tex]Therefore, the formula to use now is
[tex]p(A\cup B)=p(A)+p(B)[/tex]For Anna
Anna can join either team A or team B.
We calculate the probability
[tex]\begin{gathered} p(A\cup B)=p(A)+p(B) \\ p(A\cup B)=0.2+0.45 \\ p(A\cup B)=0.65 \end{gathered}[/tex]For Elina
Elina can join either team B or team C.
We calculate the probability
[tex]\begin{gathered} p(B\cup C)=p(B)+p(C) \\ p(B\cup C)=0.45+0.35 \\ p(B\cup C)=0.8 \end{gathered}[/tex]For Nancy
Nancy can join either team A or team C.
We calculate the probability
[tex]\begin{gathered} p(A\cup C)=p(A)+p(C) \\ p(A\cup C)=0.2+0.35 \\ p(A\cup C)=0.55 \end{gathered}[/tex]The one with the highest probability is most likely to win and that is
ELINA
Correct answer is Elina
