Question

Team A, B, and C are competing in a basketball tournament. The probability of team A winning is 0.2, the probability of team B winning is 0.45, and the probability of team C winning is 0.35. Anna can join either team A or team B. Elina can join either team B or team C. Nancy can join either team A or team C. Who is most likely to win?AnnaNancyElenaAll have an equal probability to win.

Answer 1

Solution

We are given that

$\begin{gathered} p(A)=0.2 \\ p(B)=0.45 \\ p(C)=0.35 \end{gathered}$

Note 1: Probability Formula To use

$p(A\cup B)=p(A)+p(B)-p(A\cap B)$

Note 2: Team A, B and C are Mutually Exclusive

$\begin{gathered} p(A)+p(B)+p(C)=0.2+0.45+0.35=1 \\ Th\text{ey are mutually exclusive} \\ A\cap B=B\cap C=A\cap C=\varnothing \\ p(A\cap B)=p(B\cap C)=p(A\cap C)=0 \end{gathered}$

Therefore, the formula to use now is

$p(A\cup B)=p(A)+p(B)$

For Anna

Anna can join either team A or team B.

We calculate the probability

$\begin{gathered} p(A\cup B)=p(A)+p(B) \\ p(A\cup B)=0.2+0.45 \\ p(A\cup B)=0.65 \end{gathered}$

For Elina

Elina can join either team B or team C.

We calculate the probability

$\begin{gathered} p(B\cup C)=p(B)+p(C) \\ p(B\cup C)=0.45+0.35 \\ p(B\cup C)=0.8 \end{gathered}$

For Nancy

Nancy can join either team A or team C.

We calculate the probability

$\begin{gathered} p(A\cup C)=p(A)+p(C) \\ p(A\cup C)=0.2+0.35 \\ p(A\cup C)=0.55 \end{gathered}$

The one with the highest probability is most likely to win and that is

ELINA

Correct answer is Elina