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jerryguo2282769

Answer:

a. n^2-2n+1/n squared

b. 9 green counters

Step-by-step explanation:

just as you role 3 dices and the chances are MULTIPLIED, the same goes for this problem.

n-1/n times n-1/n would create the probability of getting 2 green counters randomly from the bag's possibility.

n-1/n times n-1/n can be said as (n-1) squared or in simple terms:

n^2 -2n+1 or in words, n squared -2n +1.

divided by n squared is the fraction probability in simplest form:

n^2-2n+1/n squared

the next question is actually much easier:

0.9 is 90% in percentages, this can be detrimental to your grade if your teachers asked you for the decimal, not percentage of it.

for 1 every yellow counter, there is 9 green counters. so in the bag, there is one yellow counter meaning there are 9 green counters in the bag (after finishing the first question)

Part (a)

Answer:  (n-2)/n

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Explanation:

There are n-1 green counters to start out of n counters total.

The probability of selecting a green counter is (n-1)/n

The probability of selecting a second green counter is (n-2)/(n-1) since we're not putting the first counter back.

Multiplying the two fractions leads to (n-2)/n

Note how the (n-1) terms cancel.

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Part (b)

Answer:  19 green counters

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Explanation:

We set the result of part (a) equal to 0.9 and solve for n

(n-2)/n = 0.9

(n-2)/n = 9/10

10(n-2) = 9n ... cross multiply

10n-20 = 9n

10n-9n = 20

n = 20

We have 20 counters in the bag. One counter is yellow and the remaining n-1 = 20-1 = 19 are green.

The probability of getting one green counter is 19/20

The probability of a second green counter is 18/19

The probability of two green counters is (19/20)*(18/19) = 18/20 = 9/10 = 0.9

This helps confirm the answer.

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