There are n sweets in a bag. 6 are orange. A random sweet gets eaten, and then a second one. The probability that both sweets are orange is 1/3. Find n.

The probability that the first sweet is orange is 6/n, as 6 of the n number of sweets are orange. After an orange sweet is eaten, there are 5 orange sweets left and n-1 total sweets in the bag. This means the probability that the second sweet is orange is 5/(n-1). The probability of both sweets being orange is therefore 6/n * 5/(n-1). This can be simplified to 30/(n^2 - n), which is equal to 1/3 (from the question). The next step is to set n^2 - n to be equal to something. This can be done by manipulating 1/3. The value of this cannot be changed, as this would change the equation. However, it can be multiplied by one and the value remains the same. To match the 30 on the top of or other equation, 1 must be multiplied by 30. However we must multiply by 30/30 as this is equal to one. This gives us 1/3 * 30/30 = 30/90, which is also equal to 30/(n^2 -2).As the numerators for both fractions are equal to each other, the denominators must also be the same. Therefore we can derive the equation n^2 - n = 90. The only thing left to do now is to rearrange the equation to find n. 90 can be subtracted from both sides to give n^2 - n - 90 = 0. This should be relatively easy to factorise into (n-10)(n+9) = 0. This gives values for n of 10 and -9. However there cannot be -9 sweets in the bag, so n must be equal to 10.