**Problem:**

A bag contains (x) one rupee coins and (y) 50 paise coins. One coin is taken from the bag and put away. If a coin is now taken at random from the bag, what is the probability that it is a one rupee coin?

**Answers:**

Case I: Let the first coin removed be one rupee coin One rupee coins left = (x – 1) Fifty paise coins left = y. Probability of getting a one rupee coin in the first and second draw = x/(x + y) × (x – 1)/(x – 1 + y)

Case II: Let the first coin removed be fifty paise coin One rupee coins left = x Fifty paise coins left = y – 1. Probability of getting a fifty paise coin in the first and one rupee coin in second draw

= y / (x + y) × x / (x + y – 1)

Total probability = sum of these two = x/(x + y) [after simplification].

it doesn’t make any differnce if we take out all the coins . the probablity will remain same , because we dont know about the withdrawled coin….:)

correct!

Elaborate, please

I think the answer is incorrect because we don’t have to consider whether the first draw was 1rupee coin or 50paise coin. Answer should be (2x-1)/(x+y-1)

You may not be correct.. can describe your answer

This answer seems wrong . consider x>y . then 2x-1/x+y-1 will be greater than 1 . probability should be <=1 ,

Can someone elaborate the answer? Thank you.