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=$\left(\frac{13}{30}\right)$?
 A: $P\left(R_{1}R_{2}R_{3}\right)=P\left(R_{1}\right)P\left(R_{2}\mid R_{1}\right)P\left(R_{3}\mid R_{1}R_{2}\right)=\frac{6}{10}\frac{5}{9}\frac{4}{8}=\frac{1}{6}$
$P\left(R_{1}R_{2}^{c}R_{3}\right)+P\left(R_{1}R_{2}R_{3}^{c}\right)+P\left(R_{1}R_{2}^{c}R_{3}^{c}\right)=P\left(R_{1}\right)-P\left(R_{1}R_{2}R_{3}\right)=\frac{6}{10}-\frac{1}{6}=\frac{13}{30}$
A: The probability of first marker red is indeed $\frac35$.  However, the probability you presented is probability of the first marker red AND not all markers red.  Thus the probability of all markers red got subtracted off.
A: The probability of choosing 3 consecutive red markers is
6C3 / 10C3 (no. of favourable outcomes= 6C3, total no of outcomes= 10C3)
This evaluates to (1/6).
If you get first marker red, the other markers may or may not be red. If you say they are not all red, you reduce the number of favourable outcomes and decrease the probability (13/30 < 6/10).
For the second part of the problem.
Probability of first marker being red=6/10.
Probability of all three markers being red = 1/6
Probability of first marker being red, but not picking 3 consecutive red markers is: 6/10 - 1/6
