# Jee Advanced 2015 paper 2; code 8; q 60.

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 $\frac{4-1}{4+6-1}=\frac 1 3.$ Note by Vishnu C
4 years, 5 months ago

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- 4 years, 5 months ago

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

- 4 years, 5 months ago

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.

- 4 years, 5 months ago

No, you are slightly incorrect.

- 4 years, 5 months ago

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

- 4 years, 5 months ago

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

- 4 years, 5 months ago

Wait a minute, they said that the probability of picking up a red ball after the transfer is $\frac 1 3$. Not the entire probability including the transfer.

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

- 4 years, 5 months ago

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

- 4 years, 5 months ago

Okay, in the case where the red ball is transferred, you have in option

(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.

- 4 years, 5 months ago

Exactly.

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

- 4 years, 5 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 : )

- 4 years, 5 months ago

I think so. But then again, my thinking it's corrrect doesn't mean anything

- 4 years, 5 months ago

No. It is $P \left( \frac{red}{red \space transferred} \right).P(red \space transferred) + P \left( \frac{red}{black \space transferred} \right).P(black \space transferred)$

That's why you have to multiply.

- 4 years, 5 months ago

Looks like we have a bit of a semantics issue.

- 4 years, 5 months ago

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

- 4 years, 5 months ago

I'm getting very less marks.

- 4 years, 5 months ago

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

- 4 years, 5 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.

- 4 years, 5 months ago

You're right. It might have been slightly ambiguous.

- 4 years, 5 months ago

You ain't in Math Stack Exchange, bro! :P For typing math here, replace $\\cdots\$ with $\backslash(\cdots\backslash)$ and $\\\cdots \\$ with $\backslash[\cdots \backslash]$. The rest is the same. Also, just for the record, the \tag feature doesn't work properly here.

The MathJax formatting sucks in Brilliant. It's way better on Stack Exchange or AoPS.

- 4 years, 5 months ago

So what do you think about the problem? Any ideas?

- 4 years, 5 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. :)

- 4 years, 5 months ago

Thanks :P

- 4 years, 5 months ago

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

- 4 years, 5 months ago

- 4 years, 5 months ago

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

- 4 years, 5 months ago

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

- 4 years, 5 months ago

Whoops! Meant to say no of red balls.

- 4 years, 5 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)

- 4 years, 5 months ago

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.

- 4 years, 5 months ago

I think you haven't read the question properly.

... 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 1/3

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

- 4 years, 5 months ago

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

- 4 years, 5 months ago

Oh sorry you're rright

- 4 years, 5 months ago

Write a comment or ask a question...

- 4 years, 3 months ago