Question

a bag contains six apples. four of the apples are good and two of the apples are rotten. an apple is selected at random and is removed from the bag. then a second apple is randomly selected from the apples remaining in the bag. find the probability that:

a.) both of the apples are rotten.

b.) only one of the two apples is good

c.) the first apple is rotten given that the second apple is good

Answer 1

The process of selection is called picking without replacement.

Total In the bag = 4 good + 2 rotten = 6

In this kind of case, we calculate individual probability for each pick, using Probability = Favorable Outcomes / Total Outcomes

For every subsequent pick, the total outcomes keeps reducing by 1 and so will the favorable outcomes depending on what is being picked in the second rounds and after.

_________________________________

(a) P( 2 apples that are picked, both are rotten)

P(Rotten apple in the first pick) = 2 / 6 = 1 / 3

P(Rotten apple in the second pick) = 1 / 5 (since 1 rotten apple is gone from the first pick)

Therefore the Required probability = (1 / 3) * (1 / 5) = 1 / 15 = 0.0667

__________________________________

(b) P( 2 apples that are picked, 1 is good and one is rotten)

There are 2 ways of picking this = (Rotten first, then good) or (Good first, then rotten)

Case 1:

P(Rotten apple in the first pick) = 2 / 6 = 1 / 3

P(Good apple in the second pick) = 4 / 5

Therefore the Required probability = (1 / 3) * (4 / 5) = 4 / 15

Case 2:

P(Good apple in the first pick) = 4 / 6 = 2 / 3

P(Rotten apple in the second pick) = 2 / 5

Therefore the Required probability = (2 / 3) * (2 / 5) = 4 / 15

Therefore the required probability = 4/15 + 4/15 = 8 / 15 = 0.5333

__________________________________

(c) P(The First Rotten and Second Good) = P1 = (1/3) * (4/5) = 4/15

P(Both are Good) = P2 = (4/6) * (3/5) = 12/30 = 6/15

By Bayes Theorem, P(A given B) = P(A / B) = P(A nd B) / P(B)

Therefore P(First is rotten given the second is good) = P(First Rotten and then Good) / P(Good) = P1 / (P1 + P2)

P(Good) = P(Rotten first then good) + P(Good first and then Good) = P1 + P2 = 4/15 + 6/15 = 10/15

Therefore P(First is rotten given the second is good) = (4/15) / (10/15) = 4 / 10 = 2 / 5 = 0.4

__________________________________________

_______________________________________