Question

Suppose children like three types of chocolates Perk, Munch, and 5Star. If they
are asked to choose to pick chocolate they have their own preference, one-sixth
of children population preference is Perk>Munch>5star. One-sixth of children
population preference is Munch>5star>Perk. Similarly remaining four-sixths
children preferences follows as per above combinations. If you met a random
child and give him chance to pick a chocolate between Munch and Perk. He
picked Munch. Now you offer Munch and 5star, what is the probability that he
chooses again Munch?​

Answer 1

Given:

One-sixth of children population preference is Perk>Munch>5star.  

One-sixth of children population preference is Munch>5star>Perk.  

Similarly remaining four-sixths children preferences follow as per the above combinations.

To find:

What is the probability that he chooses again Munch?​

Solution:

From given, we have,

One-sixth of children population preference is Perk>Munch>5star.  

One-sixth of children population preference is Munch>5star>Perk.  

Possible preferences  are as follows:

5star > Perk > Munch

5star > Munch > Perk

Perk > Munch > 5star

Perk > 5star > Munch

Munch > Perk > 5star

Munch > 5star > Perk

The possibility that a child prefers Munch over Perk P(M5P), P(MP5) or P(5MP)

Then a child prefers Munch over 5star P(M5P) P(MP5)

Probability = (1/6+1/6) / (1/6+1/6+1/6) =2/3

∴ The probability that a child chooses again Munch is 2/3.