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mario62 [17]
4 years ago
7

A box contains 2 red marbles, 3 green marbles, and 3 blue marbles. if we choose a marble, then another marble without putting th

e first one back in the box, what is the probability that the first marble will be green and the second will be blue?
Mathematics
2 answers:
Bad White [126]4 years ago
7 0
There are 7 marbles, including 3 green ones, so at the start, there is a 3/7 chance of getting a green marble.
Assuming you did get a green one, there are now 6 marbles left, with 2 blue marbles, so there is a 2/6 chance of taking a blue marble.
Given that both have to happen, you must multiply each probability, hence the total probability is 3/7 x 2/6, or a 1/7 chance.
Hope this helped
Zepler [3.9K]4 years ago
4 0

Answer:

The answer is actually 9/56

Step-by-step explanation:


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\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}.

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