I saw the resonance answer key and I found that their answer to the last question (in mathematics) didn't make sense:

\(n_1\) and \(n_2\) are the number of red and black balls, respectively, in box I and \(n_3\) and \(n_4\) are the number of red and black balls, respectively, in box II. and if a randomly picked ball from box I is transferred to box II, the probability of picking a red ball from box I now becomes \(\frac 1 3\). Then the possible values of \(n_1\) and \(n_2\) are:

(a) \(n_1\)=4 and \(n_2\)=6. (b)\(n_1\)=2 and \(n_2\)=3.

(c) \(n_1\)=10 and \(n_2\)=20. (d) \(n_1\)=3 and \(n_2\)=6.

The answer key says that it is C and D. Can anyone please explain how this is possible?

I got A as my answer because \(\frac{4-1}{4+6-1}=\frac 1 3.\)

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## Comments

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TopNewest@Tanishq Varshney . @Parth Lohomi .

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You have that case also when you transfer a black ball and then pick a red one. It seems you forgot to include it.

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Did you look at the options? Did you check them to see if they match the criteria? The probability is going to be \(\frac {n_1} {n_1+n_2-1}\) or \(\frac{n_1-1}{n_1+n_2-1}\). And either of these must be \(\frac 1 3.\)

It's not really a single question. They're cheating, if you ask me. They're basically asking you 4 questions, albeit simple ones, with one single number representing it. But I'm just cribbing and I've no real problem with it.

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No, you are slightly incorrect.

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Besides, the key is from resonance. They might have gotten it wrong. That should also be considered.

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@Arpan Banerjee is right. You have to consider two cases, one when you transferred red and other when you transferred black

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(a)\(P(\frac{red}{box I})=\frac{3}{9}=\frac 1 3.\) (b) \(P(\frac{red}{box I})=\frac{1}{4}\)

(c)\(P(\frac{red}{box I})=\frac{9}{29}\) (d)\(P(\frac{red}{box I})=\frac{2}{8}\)

And in the case where the black ball is chosen you have in option

(a) \(P(\frac{red}{box I})=\frac{4}{9}\) (b)\(P(\frac{red}{box I})=\frac{2}{4}\)

(c) \(P(\frac{red}{box I})=\frac{10}{29}\) (d)\(P(\frac{red}{box I})=\frac{3}{8}\)

Is this what you're talking about? I don't understand why you have to multiply it by the probability of choosing a red ball from box I before the transfer here. The question says

after the transfer.Log in to reply

The event here is "chossing a red ball" while the transferred ball is static, not "chossing a red ball after removing a random ball."

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That's why you have to multiply.

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@Pranjal Jain How was your JEE Advanced? How much are you scoring according to Resonance answer key?

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`\tag`

feature doesn't work properly here.The MathJax formatting sucks in Brilliant. It's way better on Stack Exchange or AoPS.

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In response to Vishnu C:-

When a ball is drawn at random from box 1 and transfered to box 2, No. of red balls in box 1 is n1 - n1/(n1+n2)

_(A) And total no. of balls is n1+n2-1(B) so the required probability is A/B On substituting all the 4 options in this, only C and D turn out to be 1/3___Log in to reply

Can you type your answer in Tex? I can't read it.

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How can the number of balls in box 1 after the transfer be a fraction?

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In response to Vishnu C:- n1+n2-1 is not a fraction..

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I'm not a probability guy but it seems to me that for the probability to be \(\frac{1}{3}\) we must have \(P(A)=\frac{\text{Desired outcomes}}{\text{Sample space}}\) now for the case\(c\) the \(\text{Desired outcomes}\) are 10 since we want to calculate the probability of getting a red ball and the \(\text{Sample space}\) is \(10+20\) (As answer to the question : out of how many balls can we choose?) and hence the probability equals \(\frac{1}{3}\).

Same description should go for the case \(d\).

And sorry if I'm speaking non-sense but as I mentioned earlier,I'm not into probability stuff.

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I think you haven't read the question properly.

A ball is transferred. Then the probability becomes 1/3

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So what do you think? Is my reasoning right? How can it be C and D?

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Oh sorry you're rright

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