There are 4 baskets. The first basket has 3 apples and 4 oranges, the second one has 4 apples and 5 mangoes, the third one has 6 Mangoes and 2 bananas and the last one has 7 bananas and 2 apples. If a fruit is randomly chosen from any basket and it comes out to be an apple, then what is the probability that it was taken out from the second basket?

Answer 1

Have you done Baye's theorem, total probability and conditional probability?

If not, you need to review these topics before you can tackle this problem.

Here the assumption is not clear, but I will assume that it is equally probable to
1. pick one of the 4 baskets
2. pick any fruit of a given basket.

Using notation of condition probability:
P(A|B)= probability of event A given B has happened.

The baskets and the corresponding events will be named 1-4.
Event of choosing an apple is A.

Then
P(A|1)=3/7
P(A|2)=4/9
P(A|3)=0/8
P(A|4)=2/9

Hence event 1,2,3,4 are the partitions of the sample space.

The probability of choosing an apple, by the law of total probability
P(A)=P(A|1)P(1)+P(A|2)P(2)+P(A|3)P(3)+P(A|4)P(4)
=3/7*1/4+4/9*1/4+0/8*1/4+2/9*1/4
=23/84

By Baye's theorem,
P(2|A)=P(A|2)*P(2)/P(A)
=4/9*(1/4)/(23/84)
=28/69

Answer 2

Probability of Picking an apple from any one of the boxes should be more than picking an apple from a specific box.
But your application of Baye's Theorem for this question gives the probability the other way. can you explain that