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3 years ago

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An urn contains white and red balls. Four balls are randomly drawn from the urn in succession, with replacement. That is, after

each draw, the selected ball is returned to the urn. What is the probability that all balls drawn from the urn are red

1 answer:

Kay [80]3 years ago

5 0

Answer:

[r/(w + r)]^4

Step-by-step explanation:

The numbers of the different balls are not given, but here is the solution.

Let w = number of white balls.

Let r = number of red balls.

The total number of balls is w + r.

All drawings have the same probability since there is replacement.

Each drawing has the following probability of drawing a red ball:

p(red) = r/(w + r)

Since there are 4 drawings, and they are all independent events, the overall probability is the product of the individual probabilities.

p(4 red balls in 4 drawings) = [r/(w + r)]^4

To find an actual number, replace w and r with the correct numbers and evaluate the expression.

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$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

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can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

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Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

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$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

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