Suppose in a urn I have m identical green balls and n identical blue balls. I pick out 3 balls in random, then what is the probability that I have got 2 blue balls and 1 green ball?

I know its a very easy question and to me answer will be 1/4 ( as the balls are identical ) but then it means the probability is independent of how many blue or green balls are there. It seems counter intuitive to me, please guys comment on this.

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TopNewestwhat you guys are doing is, you are treating the balls as different ones, even though its mentioned the blue balls are identical , so are the green balls. If it was mentioned there are different blue and green balls then what you did should have been the answer. Are you trying to say that , the balls are identical doesn't make any difference to the probability ? Why is that so? please elaborate this point – Subharthi Chowdhuri · 3 years, 11 months ago

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Your conclusion that the probability is \(\frac{1}{4}\) is wrong.

Think of it this way: What is the number of ways in which you can select \(3\) balls from \(m+n\)?

Similarly you apply that for \(1\) green ball \(2\) blue balls respectively.

Thus, required probability=

\(\huge \frac{{{n}\choose{2}}{{m}\choose{1}}}{{{m+n}\choose{3}}}\) – Aditya Parson · 3 years, 11 months ago

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This problem does not have to utilize combinations at all. For case 1, let's assume that the green ball is picked first. The probability of this happening is \(\frac{m}{m+n}\). The probability we then pick a blue ball is \(\frac{n}{m+n-1}\). The probability we pick another blue ball is \(\frac{n-1}{m+n-2}\). To get our total probability, we need only multiply them together. Now, there are two other cases, where we can pick the green ball second or third. One can quickly see that the probability is the same. And since these are mutually exclusive events, we need only add them. Therefore the probability is \[\frac{3mn(n-1)}{(m+n)(m+n-1)(m+n-2)\] – Bob Krueger · 3 years, 11 months ago

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– Aditya Parson · 3 years, 11 months ago

Take note that answer will still come the same.Log in to reply

– Superman Son · 3 years, 11 months ago

rightLog in to reply

– Aditya Parson · 3 years, 11 months ago

It is not said that we draw without replacement. As such we need to use combination since we draw at random.Log in to reply

– Superman Son · 3 years, 11 months ago

dont u sleep at night bhaiLog in to reply

– Aditya Parson · 3 years, 11 months ago

Yes. I do sleep :)Log in to reply

– Superman Son · 3 years, 11 months ago

i dont sleep at night :) so i was searching for someone like meLog in to reply

I just sleep late. – Aditya Parson · 3 years, 11 months ago

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then if you are doing nC2 then it means you are treating the identical balls as different ones. Why ? suppose if i ask you, there are n identical blue balls and m identical green balls, then in how many ways you can choose 2 blue balls and 1 green ball. It must not be (nC2*mC1) as that would mean the balls are different. So the number of choice is only 1. – Subharthi Chowdhuri · 3 years, 11 months ago

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– Aditya Parson · 3 years, 11 months ago

No. Suppose a bag contains \(10\) coins of the value \(5\) and \(20\) of the value \(1\). then what is the probability of getting a 5 rupee coin?(Obviously the 5 valued coins are identical to each other and so are the 1 valued coins).Log in to reply

– Subharthi Chowdhuri · 3 years, 11 months ago

you mean you picked up 1 coin then what is the probability that coin is a 5 rupee coin? if that is the question then a single coin can be either 1 rupee coin or 5 rupee coin. hence the answer will be 1/2. if the coins were different then you could have said (10C1)/(30C1)=1/3Log in to reply