Question

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

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

Hope it helped ! 
 Mark as brainliest ! :-)