There are n urns, each of these contain...

MathematicsBayes' TheoremJEE Advanced 2006Moderate

Comprehension Passage

There are

n

urns, each of these contain

n + 1

balls. The

i^{t h}

urn contains

i

white balls and

(n + 1 - i)

red balls. Let

u_{i}

be the event of selecting

i^{t h}

urn,

i = 1, 2, 3, \dots, n

and

w

the event of getting a white ball.

Question 1:

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

(A)

(B)

2/3

(C)

3/4

(D)

1/4

Answer:B

Question 2:

If $P (u_{i}) = c$ , (a constant) then $P (u_{n} / w) =$

(A)

\frac{2}{n + 1}

(B)

\frac{1}{n + 1}

(C)

\frac{n}{n + 1}

(D)

\frac{1}{2}

Answer:A

Question 3:

Let $P (u_{i}) = \frac{1}{n}$ , if $n$ is even and $E$ denotes the event of choosing even numbered urn, then the value of $P (w / E)$ is

(A)

\frac{n + 2}{2 n + 1}

(B)

\frac{n + 2}{2 ( n + 1 )}

(C)

\frac{n}{n + 1}

(D)

\frac{1}{n + 1}

Answer:B

Visualized Solution (Hindi)

Understanding the Setup

Total balls in each urn $= n + 1$
In $i^{t h}$ urn $u_{i}$ : White balls $= i$ , Red balls $= n + 1 - i$
Conditional probability: $P (w / u_{i}) = \frac{i}{n + 1}$

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

$P (u_{i}) \propto i ⟹ P (u_{i}) = k i$
$\sum_{i = 1}^{n} P (u_{i}) = 1 ⟹ k \sum_{i = 1}^{n} i = 1$
$k \frac{n ( n + 1 )}{2} = 1 ⟹ k = \frac{2}{n ( n + 1 )}$

Total Probability of White Ball

$P (w) = \sum_{i = 1}^{n} P (u_{i}) P (w / u_{i})$
$P (w) = \sum_{i = 1}^{n} \frac{2 i}{n ( n + 1 )} \cdot \frac{i}{n + 1}$
$P (w) = \frac{2}{n ( n + 1 ) ^{2}} \sum_{i = 1}^{n} i^{2}$

Evaluating the Summation

Using $\sum i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ :
$P (w) = \frac{2}{n ( n + 1 ) ^{2}} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
$P (w) = \frac{2 n + 1}{3 ( n + 1 )}$

Limit as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n \to \infty

$lim_{n \to \infty} P (w) = lim_{n \to \infty} \frac{2 n + 1}{3 ( n + 1 )}$
Divide by $n$ : $lim_{n \to \infty} \frac{2 + 1/ n}{3 + 3/ n}$
Result: $\frac{2 + 0}{3 + 0} = \frac{2}{3}$

Part 2: Constant Probability

If $P (u_{i}) = c$ , then $n c = 1 ⟹ c = 1/ n$
$P (w) = \sum_{i = 1}^{n} \frac{1}{n} \cdot \frac{i}{n + 1} = \frac{1}{n ( n + 1 )} \sum_{i = 1}^{n} i$
$P (w) = \frac{1}{n ( n + 1 )} \cdot \frac{n ( n + 1 )}{2} = \frac{1}{2}$

Applying Bayes' Theorem

Using Bayes' Theorem: $P (u_{n} / w) = \frac{P ( w / u _{n} ) P ( u _{n} )}{P ( w )}$
$P (u_{n} / w) = \frac{\frac{n}{n + 1} \cdot \frac{1}{n}}{1/2}$
$P (u_{n} / w) = \frac{1/ ( n + 1 )}{1/2} = \frac{2}{n + 1}$

Part 3: Even Numbered Urns

Event $E = {u_{2}, u_{4}, \dots, u_{n}}$
$P (E) = \frac{n /2}{n} = \frac{1}{2}$
Need to find $P (w / E) = \frac{P ( w \cap E )}{P ( E )}$

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (w \cap E)

$P (w \cap E) = \sum_{j = 1}^{n /2} P (u_{2 j}) P (w / u_{2 j})$
$P (w \cap E) = \sum_{j = 1}^{n /2} \frac{1}{n} \cdot \frac{2 j}{n + 1} = \frac{2}{n ( n + 1 )} \sum_{j = 1}^{n /2} j$
$P (w \cap E) = \frac{2}{n ( n + 1 )} \cdot \frac{\frac{n}{2} ( \frac{n}{2} + 1 )}{2} = \frac{n + 2}{4 ( n + 1 )}$

Final Ratio for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (w / E)

$P (w / E) = \frac{P ( w \cap E )}{P ( E )}$
$P (w / E) = \frac{\frac{n + 2}{4 ( n + 1 )}}{1/2}$
$P (w / E) = \frac{n + 2}{2 ( n + 1 )}$

Summary and Takeaways

Key Takeaway: Total Probability involves summing over all possible mutually exclusive cases.
Key Takeaway: Bayes' Theorem allows us to reverse the condition once the outcome is known.
Next Challenge: What if $P (u_{i}) \propto i^{2}$ ? How would the limit change?

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B_{3}

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(B)

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(C)

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(D)

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(ui​)∝i, where i=1,2,3,…,n, then limn→∞​P(w)=

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P(ui​)=n1​, if n is even and E denotes the event of choosing even numbered urn, then the value of P(w/E) is

Visualized Solution (Hindi)

Understanding the Setup

Finding the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Total Probability of White Ball

Evaluating the Summation

Limit as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n→∞

Part 2: Constant Probability

Applying Bayes' Theorem

Part 3: Even Numbered Urns

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(w∩E)

Final Ratio for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(w/E)

Summary and Takeaways

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