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.

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

#Combinatorics #JEEAdvanced2015 #Paper2

Easy Math Editor

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Comments

@Tanishq Varshney . @Parth Lohomi .

vishnu c - 6 years 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.

Arpan Banerjee - 6 years 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.

vishnu c - 6 years ago

No, you are slightly incorrect.

Arpan Banerjee - 6 years ago

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

vishnu c - 6 years ago

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

Arpan Banerjee - 6 years ago

@Arpan Banerjee – 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.

vishnu c - 6 years ago

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

Pranjal Jain - 6 years ago

@Pranjal Jain – 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) $P(\frac{red}{box I})=\frac{1}{4}$

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

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

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 - 6 years ago

@Vishnu C – 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."

Siddhartha Srivastava - 6 years ago

@Siddhartha Srivastava – 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 : )

vishnu c - 6 years ago

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

Siddhartha Srivastava - 6 years ago

@Vishnu C – 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.

Arpan Banerjee - 6 years ago

@Arpan Banerjee – Looks like we have a bit of a semantics issue.

vishnu c - 6 years ago

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

Arpan Banerjee - 6 years ago

@Arpan Banerjee – I'm getting very less marks.

Pranjal Jain - 6 years ago

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

Arpan Banerjee - 6 years ago

@Arpan Banerjee – 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.

vishnu c - 6 years ago

@Vishnu C – You're right. It might have been slightly ambiguous.

Arpan Banerjee - 6 years ago

@Arpan Banerjee – 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.

Prasun Biswas - 6 years ago

@Prasun Biswas – So what do you think about the problem? Any ideas?

vishnu c - 6 years ago

@Vishnu C – 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. :)

Prasun Biswas - 6 years ago

@Prasun Biswas – Thanks :P

Arpan Banerjee - 6 years 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

Kaushik Bhat - 6 years ago

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

vishnu c - 6 years ago

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

vishnu c - 6 years ago

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

Kaushik Bhat - 6 years ago

@Kaushik Bhat – Whoops! Meant to say no of red balls.

vishnu c - 6 years ago

@Vishnu C – 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)

Kaushik Bhat - 6 years 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.

Arian Tashakkor - 6 years 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

Siddhartha Srivastava - 6 years ago

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

vishnu c - 6 years ago

Oh sorry you're rright

Arian Tashakkor - 6 years ago

Write a comment or ask a question...

debjani pal - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$