Probability Problem with $10$ players being put on two teams Could somebody please check my answer to this problem?
Thanks
Bob
Problem:
$10$ kids are randomly grouped into an A team with five kids and a B team with five kids. Each
grouping is equally likely. There are three kids in the group, Alex and his two best friends Jose
and Carl. What is the probability that Alex ends up on the same team with at least one of his two
best friends?
Answer:
Let $p_{a}$, $p_j$, and $p_c$ be the probabilities that Alex is on team A, Jose is on team A, and Carl is on team A. Let $p$ by the probability that we seek.
\begin{eqnarray*}
p &=& 2 ( p_a p_j + p_a * p_c  - p_a p_j p_c ) \\
p_a &=& \frac{1}{2} \\
p_j &=& \frac{1}{2} \\
p_c &=& \frac{1}{2} \\
p &=& =
 2 ( \frac{1}{2} \Big( \frac{1}{2} \Big) + \frac{1}{2} \Big( \frac{1}{2} \Big) -
  \frac{1}{2} \Big( \frac{1}{2} \Big) \Big( \frac{1}{2} \Big) \\
p &=& 2 ( \frac{1}{4} + \frac{1}{4} - \frac{1}{8} ) = 2 ( \frac{1}{2} - \frac{1}{8} ) \\
p &=& \frac{3}{4} \\
\end{eqnarray*}

Visualize $5$ slots each for team A and team B, viz $\;\;\boxed A\boxed A\boxed A\boxed A \boxed A\quad \boxed B\boxed B\boxed B\boxed B\boxed B$.  
Alex can be on any team, occupying one slot.
P(Jose and Carl are both on the other team ) $= \dfrac59\cdot\dfrac48 = \dfrac5{18}$
Hence P(at least one of them is on Alex's team) $= 1 - \dfrac5{18} = \dfrac{13}{18}$
