Let n1 and n2 be the number of red and...

MathematicsBayes' TheoremJEE Advanced 2015Moderate

Comprehension Passage

Let

n_{1}

and

n_{2}

be the number of red and black balls, respectively, in box I. Let

n_{3}

and

n_{4}

be the number of red and black balls, respectively, in box II.

Question 1:

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 1/3, then the correct option(s) with the possible values of $n_{1}, n_{2}, n_{3}$ and $n_{4}$ is(are)

(A)

n_{1} = 3, n_{2} = 3, n_{3} = 5, n_{4} = 15

(B)

n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50

(C)

n_{1} = 8, n_{2} = 6, n_{3} = 5, n_{4} = 20

(D)

n_{1} = 6, n_{2} = 12, n_{3} = 5, n_{4} = 20

Answer:A, B

Question 2:

A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball from box I, after this transfer, is 1/3, then the correct option(s) with the possible values of $n_{1}$ and $n_{2}$ is(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

Answer:C, D

Visualized Solution (English)

Visualizing the Setup

Box I: $n_{1}$ Red, $n_{2}$ Black balls.
Box II: $n_{3}$ Red, $n_{4}$ Black balls.
Event $R$ : Drawing a Red ball.
Given: $P (Box II ∣ R) = \frac{1}{3}$ .

Applying Bayes' Theorem

Let $E_{1}, E_{2}$ be the events of selecting Box I and Box II.
$P (E_{1}) = P (E_{2}) = \frac{1}{2}$
$P (R ∣ E_{1}) = \frac{n _{1}}{n _{1} + n _{2}}$ and $P (R ∣ E_{2}) = \frac{n _{3}}{n _{3} + n _{4}}$
Bayes' Theorem: $P (E_{2} ∣ R) = \frac{P ( R ∣ E _{2} ) P ( E _{2} )}{P ( R ∣ E _{1} ) P ( E _{1} ) + P ( R ∣ E _{2} ) P ( E _{2} )}$

Simplifying the Ratio

Substitute $P (E_{2} ∣ R) = \frac{1}{3}$ :
$\frac{1}{3} = \frac{\frac{n _{3}}{n _{3} + n _{4}}}{\frac{n _{1}}{n _{1} + n _{2}} + \frac{n _{3}}{n _{3} + n _{4}}}$
Rearranging: $\frac{n _{1}}{n _{1} + n _{2}} + \frac{n _{3}}{n _{3} + n _{4}} = 3 (\frac{n _{3}}{n _{3} + n _{4}})$
Final Condition: $\frac{n _{1}}{n _{1} + n _{2}} = 2 (\frac{n _{3}}{n _{3} + n _{4}})$

Verifying Options for Q1

Option (a): $\frac{3}{3 + 3} = \frac{1}{2}$ ; $2 \times \frac{5}{5 + 15} = 2 \times \frac{1}{4} = \frac{1}{2}$ . Correct.
Option (b): $\frac{3}{3 + 6} = \frac{1}{3}$ ; $2 \times \frac{10}{10 + 50} = 2 \times \frac{1}{6} = \frac{1}{3}$ . Correct.
Option (c): $\frac{8}{14} \neq = 2 \times \frac{5}{25}$ . Incorrect.
Option (d): $\frac{6}{18} \neq = 2 \times \frac{5}{25}$ . Incorrect.

Transfer Scenario

A ball is transferred from Box I to Box II.
New Goal: Find $P (Red from Box I after transfer) = \frac{1}{3}$ .
Two cases for the transferred ball: Red or Black.

Total Probability Setup

Let $T_{R}$ be 'Red transferred' and $T_{B}$ be 'Black transferred'.
$P (R_{a f t er}) = P (R ∣ T_{R}) P (T_{R}) + P (R ∣ T_{B}) P (T_{B})$
$P (R_{a f t er}) = (\frac{n _{1} - 1}{n _{1} + n _{2} - 1}) (\frac{n _{1}}{n _{1} + n _{2}}) + (\frac{n _{1}}{n _{1} + n _{2} - 1}) (\frac{n _{2}}{n _{1} + n _{2}})$

The 'Aha!' Simplification

Factor out $n_{1}$ : $\frac{n _{1} ( n _{1} - 1 + n _{2} )}{( n _{1} + n _{2} - 1 ) ( n _{1} + n _{2} )}$
Cancel $(n_{1} + n_{2} - 1)$ : $P (R_{a f t er}) = \frac{n _{1}}{n _{1} + n _{2}}$
Given condition: $\frac{n _{1}}{n _{1} + n _{2}} = \frac{1}{3}$

Final Conclusion

Check $\frac{n _{1}}{n _{1} + n _{2}} = \frac{1}{3}$ :
Option (c): $\frac{10}{10 + 20} = \frac{10}{30} = \frac{1}{3}$ . Correct.
Option (d): $\frac{3}{3 + 6} = \frac{3}{9} = \frac{1}{3}$ . Correct.
Final Answers: Q1: (a, b); Q2: (c, d)

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There are

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w

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If $P (u_{i}) \propto i$ , where $i = 1, 2, 3, \dots, n$ , then $lim_{n \to \infty} P (w) =$

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(A)

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(B)

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(D)

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Comprehension Passage

Visualized Solution (English)

Visualizing the Setup

Applying Bayes' Theorem

Simplifying the Ratio

Verifying Options for Q1

Transfer Scenario

Total Probability Setup

The 'Aha!' Simplification

Final Conclusion

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