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svetoff [14.1K]
3 years ago
6

A student takes a multiple choice test. Each question has N answers. If the student knows the answer to a question, the student

gives the right answer, and otherwise guesses uniformly and at random. The student knows the answer to 70% of the questions. Write K for the event a student knows the answer to a question and R for the event the student answers the question correctly. (a) What is P(K)? (b) What is P(RIK)? (c) What is P(KR), as a function of N? (d) What values of N will ensure that P(K|R) >99%?
Mathematics
1 answer:
alexgriva [62]3 years ago
8 0

Answer:

Step-by-step explanation:

a) P(K) = 0.7

b) P(R|K) = 1

c) P(K|R) = P(KnR) /P(R)

= P(K) x P(R|K) / ( P(K) x P(R|K) + P(K') P(R|K')

= 0.7 x 1 / ( 0.7 x 1 + (1-0.7) x 1/N)

= 0.7/(0.7 + 0.3/N)

d) P(K|R) > 0.99

0.7/(0.7 + 0.3/N) > 0.99

N > 42.86

N = 43 will be the value of N that will ensure that P(K|R) >99%

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There are 6 red marbles 4 blue marbles to yellow marbles. Sabrina picks 2 marbles without putting the first one back what is the
Yanka [14]

Answer:

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Step-by-step explanation:

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We want to find the probability that the first pick is not yellow (so it can be either blue or red), and the second pick is a blue marble.

Now there are two cases.

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q = 4/11

The joint probability is equal to the product of the individual probabilities, then:

P₁ = p*q = (6/12)*(4/11) = 2/11

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For the first draw, the probability is:

p = 4/12

And for the second draw, now there are 3 blue marbles and 11 total marbles, then the probability is:

q = 3/11

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3 years ago
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3 years ago
Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units. What would be the perimeters of the
Dmitriy789 [7]

Answer:

The perimeter of other rectangles are 50, 28, 22 units

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Step-by-step explanation:

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Proof -

We know that

Are of Rectangle = Length × Breadth

And Perimeter of Rectangle = 2 (Length + Breadth )

Now,

Given that,

Orbet used 24 square tiles

And perimeter of the rectangle made = 20

So,

Possible Length of rectangle = 6

Possible breadth of Rectangle = 4

Or Vice-versa.

Total number of Rectangles possible = 4

Possibilities are -  1 × 24, 2 × 12, 3 × 8, 4 × 6

Case I :

Length of rectangle = 1

Breadth of rectangle = 24

∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units

Area of rectangle = 1 × 24 = 24 unit²

Case II :

Length of rectangle = 2

Breadth of rectangle = 12

∴ Perimeter of rectangle = 2(2 + 12) = 2(14) = 28 units

Area of rectangle = 2 × 12 = 24 unit²

Case III :

Length of rectangle = 3

Breadth of rectangle = 8

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Area of rectangle = 3 × 8 = 24 unit²

Case IV :

Length of rectangle = 4

Breadth of rectangle = 6

∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units

Area of rectangle = 4 × 6 = 24 unit²

∴ we get

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

5 0
3 years ago
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