Committee of 50 senators. Each state of the 50 in the U.S. has 2 senators. What is the probability that in a random committee of 50 senators.
California is represented?  
I got $\dfrac{_2Cn_1 \times {_{98}Cn_{49}}}{_{100}Cn_{50}}$
Is that correct? If not where is my mistake?.
Thanks.
 A: There are $\binom{98}{50}$ committees that do not represent California, hence the answer is given by:
$$ 1-\frac{\binom{98}{50}}{\binom{100}{50}}=1-\frac{50\cdot 49}{100\cdot 99}=\frac{149}{198}\approx\color{red}{75,25\%}. $$
A: I'm not sure why you have $_2C_1$. I am a little bit inclined to suspect it has to do with choosing one senator from California.  But in fact it is possible that both are chosen.  Not taking into account the possibility that both are chosen is an error.
One way to proceed is as in Jack d'Aurizio's answer.  Here's another:
The probability that Senator $A$ is chosen is $1/2$.  One way to see that is that the number of times Senator A is chosen is either $0$ or $1$, and the probability that he is chosen is the expected value of that number, and the sum of the expected values over all $100$ senators is the expected value of the sum, and that is $50$.
But now what is the conditional probability that Senator B is not chosen, given that Senator A is not chosen?  Given that Senator A is not chosen, the problem becomes that of choosing $50$ out of $99$, and an argument like that in the paragraph above then shows that the conditional probability that Senator B is not chosen is $49/99$.
So:
\begin{align}
& \Pr(\text{Senators A and B are both excluded}) \\[10pt]
= {} & \Pr(\text{Senator A is excluded}) \times \Pr(\text{Senator B is excluded} \mid \text{Senator A is excluded}) \\[10pt]
= {} & \frac 1 2 \times \frac{49}{99} = \frac{49}{198} = 0.2474747\ldots
\end{align}
So the probability that California is represented is
$$
0.7525252\ldots
$$
