Q:

Probability of winning in competition

Fifty teams compete in a student programming competition. It has been observed that 60% of the teams use the programming language C while the others use C++, and experience has shown that teams who program in C are twice as likely to win as those who use C++. Furthermore, ten teams who use C++ include a graduate student, while only four of those who use C include a graduate student.

What is the probability that the winning team programs in C?
What is the probability that the winning team programs in C and includes a graduate student?
What is the probability that the winning team includes a graduate student?
Given that the winning team includes a graduate student, what is the probability that that team programmed in C?

My answers are: 

Dividing probability of team C won by the sum of probabilities team C  $\frac {6/10*2/3}{6/10*2/3+4/10*1/3}=\frac 34$ 
C team has 4 graduate students so: 
$\frac 34 * \frac 4{30}=\frac 1{10}$
C++ team has 10 graduate students so:
$\frac 34 * \frac 4{30} + \frac 14 * \frac {10}{20}=\frac 9{40}$
I consider both teams with graduate students and look only for the C team: $\frac{6/10*2/3*4/30}{6/10*2/3*4/30+4/10*1/3*10/20}=\frac 49$

Is it correct ? Especially the 4. Thanks in advance.

A:

This all looks good to me. Another way to look at part $4$:
The chance a team programming in C wins the competition and has a grad student is $(4/30)*(3/4) = {1 \over 10}$ since there's a $3/4$ chance the winning team programs in $C$ and $4$ of the $30$ C teams have a grad student. Similarly, the chance a team programming in C++ wins and has a grad student is $(10/20)*(1/4) = {1 \over 8}$. Hence the chance a $C$ team wins given that one of these two scenarios is occurring is 
$${{1 \over 10} \over {1 \over 10} + {1 \over 8}} = {4 \over 9}$$

