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. If the first marker she picks is red but not all 3 markers
are red, she will win a total of $100. Under any other outcome, she will win $0.
What is the expected value of Jessica's winnings?
Round your answer to the nearest cent.

The probability of Jessica picking 3 consecutive red markers is: (1/6)

The probability of Jessica's first marker being red, but not picking 3 consecutive red markers is:

(3/5)−(1/6)=(13/30)

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)=(6/10) can any explain me the probability of Jessica's first marker being red=(13/30)?

Step-by-step explanation:

goog may be able to you find that answer

she drove 134 miles

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