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uysha [10]
3 years ago
13

Jessica is playing a game where there are 4 blue markers and 6 red markers in a box. She is going to pick 3

Mathematics
1 answer:
Andru [333]3 years ago
5 0

Answer:

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:

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Step-by-step explanation:

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7 0
3 years ago
Prove that is here<br><img src="https://tex.z-dn.net/?f=1%20-%20cos%20%7B2%7Da%20%5Cdiv%201%20-%20sin%20a%7B2%7D%20%20%3D%20tan%
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\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}

<h3>LHS</h3>

\boxed{\sf \dfrac{cosA}{sinA}=cotA}

\\ \sf\longmapsto \dfrac{1-cos2A}{1-sin2A}

\\ \sf\longmapsto 1-cot2A

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