Question

# 1. (2+6 marks) (a) State Bayes’ Theorem. (b) There are n boxes, the 1st one containing...

1. (2+6 marks) (a) State Bayes’ Theorem.

(b) There are n boxes, the 1st one containing 1 glass, the 2nd box containing 2 glasses, the 3rd box containing 3 glasses, . . . , kth box containing k glasses, . . . nth box containing n glasses. In each box every glass is broken with probability 1/2 independently of all other glasses. A contestant chooses a box uniformly at random. Given that he found no broken glasses in this box, find the probability that he choose the 1st box. Simplify your answer.

Given that the person found no broken glasses in the box, the probability that he choose 1st box is computed using Bayes theorem here as:

= Probability that the 1st box is chosen * Probability of no broken glass in the 1st box / Probability of no broken glass in the box chosen

First using law of total probability, we have here:

Probability of no broken glass in the box chosen
= Probability that first box is chosen * Probability of no broken glass in the 1st box +
Probability that second box is chosen * Probability of no broken glass in the 2nd box + ...... n boxes   Therefore the required probability now is computed here as:  This is the required probability here.

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