profit from lottery ticket Jessica is playing a game where there are 4 blue markers and 6 red markers in a box. She is going to pick 3 markers without replacement.
If she picks all 3 red markers, she will win a total of 500 dollars. If the first marker she picks is red but not all 3 markers are red, she will win a total of 100 dollars. Under any other outcome, she will win 0 dollars. 
Solution
The probability of Jessica picking 3 consecutive red markers is: $\left(\frac16\right)$
The probability of Jessica's first marker being red, but not picking 3 consecutive red markers is:$\left(\frac35\right)-\left(\frac16\right)=\left(\frac{13}{30}\right)$

So i am bit stuck here
what i think is it shouldn't be that complex it should be as simple as 
 chance of Jessica's first marker being red=chance of getting red 1 time
i.e P(First marker being red)=$\left(\frac{6}{10}\right)$
can any explain me the probability of Jessica's first marker being red=$\left(\frac{13}{30}\right)$?

$P\left(R_{1}R_{2}R_{3}\right)=P\left(R_{1}\right)P\left(R_{2}\mid R_{1}\right)P\left(R_{3}\mid R_{1}R_{2}\right)=\frac{6}{10}\frac{5}{9}\frac{4}{8}=\frac{1}{6}$
$P\left(R_{1}R_{2}^{c}R_{3}\right)+P\left(R_{1}R_{2}R_{3}^{c}\right)+P\left(R_{1}R_{2}^{c}R_{3}^{c}\right)=P\left(R_{1}\right)-P\left(R_{1}R_{2}R_{3}\right)=\frac{6}{10}-\frac{1}{6}=\frac{13}{30}$
