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There are "n" sweets in a bag, six are orange and the rest are yellow. If you take a random sweet from the bag and eat it. Then take at random another one and eat it. The probability of eating two orange sweets is 1/3. Show that n²-n-90=0.

We have:

n= total of sweets

6= orange sweets

(6-n)=yellow sweets (We use 6-n beacuse we know that if 6 sweets are orange, the rest must be yellow, so yellow sweets= (total of sweets-orange sweets))

If the probability of geting two orange aweets is 1/3, then:

(6/n) x (5/(n-1))= 1/3

Here, 6 over n is the probability of getting an orange sweet, we use Laplace´s Law: (number of favourable cases)/(number of total cases), that would mean: number of orange sweets/ total number of sweets. So if we have already eaten an orange sweet, there are 5 orange sweets left and the total number of sweets is n-1, that is why the second fraction is 5/(n-1)

Then we get:

30/(n^2-n)= 1/3

We try to isolate the n (as it is an equation):

n^2 - n= 30x3

n^2 - n= 90

Therefore;

n^2 - n - 90=0

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There are "n" sweets in a bag, six are orange and the rest are yellow. If you take a random sweet from the bag and eat it. Then take at random another one and eat it. The probability of eating two orange sweets is 1/3. Show that n²-n-90=0.

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