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.
Dude
If 1 coin is drawn that may be re.1 or 50 paise
So when that coin is drawn out 1 coin is reduced that can be any of the re.1 or 50 paise
So
If that withdrawn coin was re.1
P1= (x-1)/(x+y-1)
If that was 50 paisa
P2=(x)/(x+y-1)
=>
P =P1*P2 = x(x-1)/(x+y-1)^2
Hi,
I can think of two almost equivalent ways for seeing why the answer is 1/51. The first one is the most straightforward one, while the second one is simpler.
1.
Let’s define the events:
1r=the 1st taken coin was a rupee, 1p= the 1st taken coin was a paise,
2r=the 2nd taken coin was a rupee, 2p= the 2nd taken coin was a paise.
The probability of getting the rupee on the second coin is then
P(1r) * P(2r | 1r) + P(1p) * P(2r | 1p)
Now, if you choose first a rupee, then the probability of getting again a rupee is zero. So P(2r | 1r) is zero and the first term is zero. For the second term:
P(1p) = 50/51 and P(2r | 1p) = 1/50. And so we got the answer multiplying those numbers.
2.
You get your hand in the bag and take out one coin. It’s the same (probability talking) if, previously of getting a coin, you touch another coin and deciding that you will not choose it. That is, the probability is the same as if we didn’t discard another coin first.