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 \(\frac{4-1}{4+6-1}=\frac 1 3.\)

## Comments

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TopNewestIn 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 – Kaushik Bhat · 1 year, 8 months ago___Log in to reply

– Vishnu C · 1 year, 8 months ago

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

– Kaushik Bhat · 1 year, 8 months ago

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

– Vishnu C · 1 year, 8 months ago

Whoops! Meant to say no of red balls.Log in to reply

– Kaushik Bhat · 1 year, 7 months ago

total no. of balls in box 1 is n1+n2-1 .............But no. of red balls are n1 initially.......... But when a ball is drawn,the probability that it is red is n1/(n1+n2)... and blue is n2/(n1+n2)... So,in total(talking of probability), 1 single ball is drawn... Hence, the no. of red balls after drawing a random ball is n1-n1/(n1+n2)Log in to reply

– Vishnu C · 1 year, 8 months ago

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

You have that case also when you transfer a black ball and then pick a red one. It seems you forgot to include it. – Arpan Banerjee · 1 year, 8 months ago

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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. – Vishnu C · 1 year, 8 months ago

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– Arpan Banerjee · 1 year, 8 months ago

No, you are slightly incorrect.Log in to reply

– Vishnu C · 1 year, 8 months ago

Help me then. I don't get it. That's why I posted the note.Log in to reply

– Arpan Banerjee · 1 year, 8 months ago

You also have to multiply with the probability of transferring that colour of ball. Overall your expression would become \[\left( \frac{n_1}{n_1+n_2} \right) \left( \frac{n_1-1}{n_1+n_2-1} \right) + \left( \frac{n_2}{n_1+n_2} \right) \left( \frac{n_1}{n_1+n_2-1} \right)\]Log in to reply

Besides, the key is from resonance. They might have gotten it wrong. That should also be considered. – Vishnu C · 1 year, 8 months ago

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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 – Pranjal Jain · 1 year, 8 months ago

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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. – Vishnu C · 1 year, 8 months agoLog in to reply

That's why you have to multiply. – Arpan Banerjee · 1 year, 8 months ago

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– Vishnu C · 1 year, 8 months ago

Looks like we have a bit of a semantics issue.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." – Siddhartha Srivastava · 1 year, 8 months ago

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– Vishnu C · 1 year, 8 months ago

So do you think I'm right? That means that resonance key is wrong. That also means that I got all the questions that I attempted in mathematics right : )Log in to reply

– Siddhartha Srivastava · 1 year, 8 months ago

I think so. But then again, my thinking it's corrrect doesn't mean anythingLog in to reply

@Pranjal Jain How was your JEE Advanced? How much are you scoring according to Resonance answer key? – Arpan Banerjee · 1 year, 8 months ago

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– Pranjal Jain · 1 year, 7 months ago

I'm getting very less marks.Log in to reply

– Arpan Banerjee · 1 year, 7 months ago

I am too. But don't worry, there will be quite a drop from last year :)Log in to reply

`\tag`

feature doesn't work properly here.The MathJax formatting sucks in Brilliant. It's way better on Stack Exchange or AoPS. – Prasun Biswas · 1 year, 8 months ago

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– Arpan Banerjee · 1 year, 8 months ago

Thanks :PLog in to reply

– Vishnu C · 1 year, 8 months ago

So what do you think about the problem? Any ideas?Log in to reply

– Prasun Biswas · 1 year, 8 months ago

Haha, no, I don't think I'll be of any help here. Combinatorics and classical geometry are few of my weak points and like Arian, I'm not much of a probability guy too. Still, I'll try that problem later in my free time if I can. :)Log in to reply

– Vishnu C · 1 year, 8 months ago

But isn't what they're asking \(P(\frac {red}{box I})\)? That's what the question sounds like to me. If they meant to ask what you understood, then I'm afraid I was wrong.Log in to reply

– Arpan Banerjee · 1 year, 8 months ago

You're right. It might have been slightly ambiguous.Log in to reply

@Tanishq Varshney . @Parth Lohomi . – Vishnu C · 1 year, 8 months ago

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Write a comment or ask a question... – Debjani Pal · 1 year, 6 months ago

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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. – Arian Tashakkor · 1 year, 8 months ago

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A ball is transferred. Then the probability becomes 1/3 – Siddhartha Srivastava · 1 year, 8 months ago

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– Arian Tashakkor · 1 year, 8 months ago

Oh sorry you're rrightLog in to reply

– Vishnu C · 1 year, 8 months ago

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