Question

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

Answer 1

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