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Probability question

2015 · Shift 2 · Q25
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  5. /2015 · Shift 2 · Q25

Probability question

2015 · Shift 2 · Q25

JEE AdvancedMathematicsProbabilityMultiple correct+4 / −1
Let n1{n_1}n1​ and n2{n_2}n2​ be the number of red and black balls, respectively, in box I{\rm I}I. Let n3{n_3}n3​ and n4{n_4}n4​ be the number of red and black balls, respectively, in box II.{\rm I}{\rm I}.II. A ball is drawn at random from box I{\rm I}I and transferred to box II.{\rm I}{\rm I}.II. If the probability of drawing a red ball from box I,{\rm I},I, after this transfer, is 13,{1 \over 3},31​, then the correct option(s) with the possible values of n1{n_1}n1​ and n2{n_2}n2​ is(are)
  1. A
    n1=4{n_1} = 4n1​=4 and n2=6{n_2} = 6n2​=6
  2. B
    n1=2{n_1} = 2n1​=2 and n2=3{n_2} = 3n2​=3
  3. C
    n1=10{n_1} = 10n1​=10 and n2=20{n_2} = 20n2​=20
  4. D
    n1=3{n_1} = 3n1​=3 and n2=6{n_2} = 6n2​=6
View written solutionFree

Correct answer: C, D

  1. Interpret the condition carefully

A ball is drawn from box I and transferred to box II.
After this transfer, the probability of drawing a red ball from box II becomes 13\frac{1}{3}31​.

Let box I initially contain: n1 red,n2 blackn_1 \text{ red},\quad n_2 \text{ black}n1​ red,n2​ black

Let box II initially contain: n3 red,n4 blackn_3 \text{ red},\quad n_4 \text{ black}n3​ red,n4​ black

Since one ball is transferred from box I to box II, there are two possibilities:

  • A red ball is transferred with probability n1n1+n2\frac{n_1}{n_1+n_2}n1​+n2​n1​​
  • A black ball is transferred with probability n2n1+n2\frac{n_2}{n_1+n_2}n1​+n2​n2​​

After transfer:

  • if red is transferred, box II has red-ball probability n3+1n3+n4+1\frac{n_3+1}{n_3+n_4+1}n3​+n4​+1n3​+1​
  • if black is transferred, box II has red-ball probability n3n3+n4+1\frac{n_3}{n_3+n_4+1}n3​+n4​+1n3​​

Hence, by total probability, the probability of drawing a red ball from box II after transfer is

+ \frac{n_2}{n_1+n_2}\cdot \frac{n_3}{n_3+n_4+1}$$ This is given to be $\frac13$. --- 2. **Simplify the expression** $$\frac{1}{n_3+n_4+1}\left(\frac{n_1(n_3+1)+n_2n_3}{n_1+n_2}\right)=\frac13$$ Now, $$n_1(n_3+1)+n_2n_3=n_1n_3+n_1+n_2n_3=(n_1+n_2)n_3+n_1$$ So, $$\frac{(n_1+n_2)n_3+n_1}{(n_1+n_2)(n_3+n_4+1)}=\frac13$$ Equivalently, $$\frac{n_3+\frac{n_1}{n_1+n_2}}{n_3+n_4+1}=\frac13$$ Rearranging, $$3\left((n_1+n_2)n_3+n_1\right)=(n_1+n_2)(n_3+n_4+1)$$ This becomes $$2(n_1+n_2)n_3+3n_1=(n_1+n_2)(n_4+1)$$ Since $n_3,n_4$ are unspecified, we need those options for which this equation can hold for some positive integers $n_3,n_4$. A more useful form is: $$\frac{n_1}{n_1+n_2}=\frac13(n_3+n_4+1)-n_3$$ The right-hand side must be compatible with integers $n_3,n_4$, so the left-hand side should allow such a form. Let us test the options through the ratio $\frac{n_1}{n_1+n_2}$. --- 3. **Evaluate each option** ### Option A: $n_1=4,\; n_2=6$ $$\frac{n_1}{n_1+n_2}=\frac{4}{10}=\frac25$$ Then the required condition becomes $$\frac{n_3+\frac25}{n_3+n_4+1}=\frac13$$ So, $$3\left(n_3+\frac25\right)=n_3+n_4+1$$ $$2n_3+\frac65=n_4+1$$ $$n_4=2n_3+\frac15$$ This is impossible since $n_4$ must be an integer. So **A is not possible**. --- ### Option B: $n_1=2,\; n_2=3$ $$\frac{n_1}{n_1+n_2}=\frac{2}{5}$$ Exactly the same fraction as in A, so again $$n_4=2n_3+\frac15$$ Impossible for integer $n_4$. So **B is not possible**. --- ### Option C: $n_1=10,\; n_2=20$ $$\frac{n_1}{n_1+n_2}=\frac{10}{30}=\frac13$$ Now, $$\frac{n_3+\frac13}{n_3+n_4+1}=\frac13$$ So, $$3\left(n_3+\frac13\right)=n_3+n_4+1$$ $$3n_3+1=n_3+n_4+1$$ $$n_4=2n_3$$ This is possible for many integer choices of $n_3,n_4$ (for example $n_3=1, n_4=2$). So **C is possible**. --- ### Option D: $n_1=3,\; n_2=6$ $$\frac{n_1}{n_1+n_2}=\frac{3}{9}=\frac13$$ Again, $$\frac{n_3+\frac13}{n_3+n_4+1}=\frac13$$ This gives $$n_4=2n_3$$ Possible for integer values of $n_3,n_4$. So **D is possible**. --- 4. **Final answer** The possible values of $n_1$ and $n_2$ are: $$\boxed{\text{C and D}}$$ --- 5. **Comparison with stored answer** Stored correct answer: **C, D** Derived answer: **C, D** So the derived answer agrees with the stored answer.
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