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Zielflug [23.3K]
3 years ago
14

Two marbles are drawn without replacement from a box with 3 white, 2 green, 2 red, and 1 blue marble. find the probability that

both marbles are white.
Mathematics
1 answer:
Kaylis [27]3 years ago
5 0
<span>3/28 exactly, or approximately 10.7% Since we're drawing without replacement, let's first calculate the probability that the 1st marble is white. There are 3 white marbles out of a total of 8 marbles. So the probability of the 1st marble being white is 3/8. Now assuming that the first marble picked is white, the probability of the second marble is 2 out of the 7 remaining marbles. So it's probability is 2/7. This means that the probability of both marbles being white is 3/8 * 2/7 = 6/56 = 3/28, or approximately 10.7%</span>
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Step-by-step explanation:

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Differentiate the second time:

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3y^{2}.\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} =  - \frac{-2x}{y^{5}}

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4 0
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