Suppose there are three bags such that
Bag A has 10 bulbs out of which 4 are defective.
Bag B has 6 bulbs out of which 1 is defective
Bag C has 8 bulbs out of which 3 are defective.
A bag is chosen at random and then a bulb is randomly chosen from the chosen bag.
(i) Find the probability that the bulbs is non-defective
(ii) If the bulb is non-defectivem, find the probability that it came from bag C?

Explanation

The tree diagram describes the process given in the problem and gives the probability of each edge of the tree. The multiplication theorem then gives us the probability of a given path of the tree as the product of probabilities of each edge of the path. For instance, the probability of choosing bag A and then a non defective bulb N from bag A is `1/3xx3/5=1/5`
(i) because there are three disjoint paths leading to a non-defective bulb N, the sum of the probabilities of these paths gives us the required probability.
`P(N)=1/3.(3)/(5)+1/3.(5)/(6)+1/3.(5)/(8)=1/3(3/5+5/6+5/8)`
`=1/3((72+100+75))/(120)=247/360`
(ii) Now we want to compute P(C/N), i.e., the conditional probability of bag C given a non defective bulb N.
Now the event of bag C and a non defective bulb N, that is, `CcapN`, can only on the bottom path.
`P(CcapN)=1/3.(5)/(8)=5/24`
By definition of conditional probability
`P(C//N)=(P(CcapN))/(P(N))=(5//24)/(247//360)=75/247`
Put other way, the conditional probability in the tree diagram is the probability of successful path divided by the probability of reduced sample space consisting of all paths leading to N.