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*}
 A: The probability that Alex ends up on the same team as at least one of his two best friends can be found by subtracting the probability that neither of his friends is on the same team from $1$.  If Alex is on a team, four of the other nine kids must be his teammates.  They can be selected in 
$$\binom{9}{4}$$
ways.  If both Jose and Carl are not on his team, then Alex's teammates must be selected from the other seven available students.  The number of such selections is 
$$\binom{7}{4}$$
Hence, the probability that neither Jose nor Carl is on the same team as Alex is 
$$\frac{\dbinom{7}{4}}{\dbinom{9}{4}} = \frac{35}{126} = \frac{5}{18}$$
Therefore, the probability that at least one of his two best friends are on the same team as Alex is 
$$1 - \frac{\dbinom{7}{4}}{\dbinom{9}{4}} = \frac{13}{18}$$
As for why our results differ, notice that once Alex has been placed on a team, the probability that Jose will be placed on the same team is $4/9$ since there are only four positions left to be filled and nine students available.  The same calculation applies to Jose.  The probability that both his friends are on the same team as Alex is $\frac{4}{9} \cdot \frac{3}{8}$.  To see this, observe that if Jose is selected to be on the same team as Alex, there are three positions left for Alex and eight people left to fill those positions.  Hence, the probability that at least one of Alex's friends is on the same team is 
$$P(A \cap J) + P(A \cap C) - P(A \cap J \cap C) = \frac{4}{9} + \frac{4}{9} - \frac{4}{9} \cdot \frac{3}{8} = \frac{13}{18}$$
A: 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) $= 1 - \dfrac5{18} = \dfrac{13}{18}$
