Question

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.

Answer #1

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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