Maths Olympiad - Probability 1

Question:

A bag contains x green and y red sweets. A sweet is selected at random and its colour noted. It is then replaced into the bag together with 10 additional sweets of the same colour. A second sweet is next randomly drawn. Find the probability that the second sweet is red.

Answer:

Total no. of sweets in the bag at the start = (x + y)

Probability that first sweet is GREEN = x / (x +y)

Probability that first sweet is RED = y / (x +y)

If the first sweet was GREEN:

Then, 10 more GREEN sweets will be added.

Therefore, total no. of GREEN sweets = (x + 10)

Therefore, total no. of sweets in the bag is now = (x + 10) + y

Therefore, the probability that the second sweet is RED = y / (x + 10 + y)

If the first sweet was RED:

Then, 10 more RED sweets will be added.
Therefore, total no. of RED sweets = (y + 10)
Therefore, total no. of sweets in the bag is now = (y + 10) + x
Therefore, the probability that the second sweet is RED = (y + 10) / (y + 10 + x)

Therefore, the probability that the second sweet is RED

= [probability of first GREEN, then RED] + [probability of first RED, then RED]

= [x / (x +y)] * [y / (x + 10 + y)] + [y / (x +y)] * [(y + 10) / (y + 10 + x)]

= y / (x + y)

*You may also use a probability tree to illustrate the solution.