Answer and Explanation:
Hannah has 6 orange sweets and some unknown number of yellow sweets to make sweets in total. Note that we don’t need the number of yellow sweets she has, but you can easily calculate once we know it will simply be − 6. Ok she takes two of them we are told that the probability that they are both orange is [tex]\frac{1}{3}[/tex]. This means that [tex]\frac{6}{n} x \frac{5}{n-1} = \frac{1}{3}[/tex] at first, she has a 6 in chance to get an orange one, in the next step both the orange pile as well as the overall pile has decreased.
Alright let’s solve this:
[tex]\frac{6}{n} x \frac{5}{n-1} = \frac{1}{3} | (3n^{2} - n)[/tex]
[tex]90 = n^{2} - n | - 90[/tex]
[tex]n^{2} - n - 90 = 0[/tex]
And hey that’s the form you are looking, so ◼
But let’s continue anyway, I just use the standard formula:
[tex]n = \frac{-(-1)+\sqrt{(-1)^{2}-4x1x(-90) } }{2x1}[/tex] = -9 V 10
Obviously, can only be 10 here. Hannah has 10 sweets 6 of the orange (and 4 yellow). When she takes the first sweet it has a [tex]\frac{6}{10} = \frac{3}{5}[/tex] chance of being orange, when she gets an orange one and then picks a second at random it has a [tex]\frac{5}{9}[/tex] chance of being orange as well we multiply the two chances to see what the probability is that they happen in succession and get [tex]\frac{1}{3}[/tex] (the 5s and the 3 and the 9 cancel each other out).
