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Alona [7]
3 years ago
8

An urn contains ten chips. An unknown number of chips are white' the others are red. We wish to test:

Mathematics
1 answer:
Olin [163]3 years ago
5 0

Answer:

Let’s denote X to be the number of white chips in the sample and E be the event that exactly half of the chips  are white. Then,

a) Find α

α = P (reject H0 | H0 is true) = P (X ≥ 2|E)  

  = P (X = 2|E) + P (X = 3|E),    

We took two case, as we can draw only only three chips with two or more white to reject H0, it means we can only take 2 white chips or 3, not more, we get solution

  =  (5C2 * 5C1)/10C3  +  (5C3 * 5C0)/10C3

  =  0.5

So,  α = 0.5

b) Find β

i) Let E1 be the event that the urn contains 6 white and 4 red chips. (As given)

β = P (accept H0 | E1) = P (X ≤ 1|E1)

  =  (6C0 * 4C3)/10C3  +  (6C1 * 4C2)/10C3

 = 1/3

 = 0.333

So, β = 0.333

i) Let E2 be the event that the urn contains 7 white and 3 red chips. (As given)

β = P (accept H0 | E2) = P (X ≤ 1|E2)

  =  (7C0 * 3C3)/10C3  +  (7C1 * 3C2)/10C3

 = 11/60

 = 0.183

So, β = 0.183

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Question:

What is the common denominator of $y+\frac{y-3}{3}$ in the complex fraction $\frac{y+\frac{y-3}{3}}{\frac{5}{y}+\frac{2}{3 y}}$

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Answer:

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Explanation:

It is given that the complex fraction $\frac{y+\frac{y-3}{3}}{\frac{5}{y}+\frac{2}{3 y}}$

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The LCM can be determined by multiplying the denominators.

Thus, we get,

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Thus, the common denominator is 3.

Hence, Option D is the correct answer.

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