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].
Hint
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….:)
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].
Hint
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….:)
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