There are "n" delicious sweets in the bag. you know that 6 of them are orange and the rest are yellow. hannah takes a random sweet from the bag and eats it. then she takes a second sweet from the bag and eats that one. you know that the probability that hannah eats two of the orange sweets is 1/3. how many sweets were in the bag before hannah ate any

At the point when Hannah takes her first sweet from the sack, there is a 6/n chance it is orange. 
This is because that there are 6 orange desserts and n desserts altogether. 
When Hannah takes out her second sweet, there is a 5/(n-1) chance that it is orange. 
This is because there are just 5 orange desserts let alone for an aggregate of n-1 desserts. 
The possibility of getting two orange desserts in succession is the main likelihood increased by the second one: 6/n x 5/n–1 
The question lets us know that the shot of Hannah getting two orange desserts is 1/3. 
So: 6/n x 5/n–1 = 1/3 
Now, rearrange this problem. 
(6x5)/n(n-1) = 1/3 
This gets to be: 
30/(n² – n) = 1/3 
Times by 3 on both sides: 
90/(n² – n) = 1 
What's more, doing likewise with (n² – n): 
So (n² – n) = 90 
Our answer is: n² – n – 90 = 0

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