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Animated Solution for Mathematics - Probability: A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

Select Answer:

Progressive Hints

Hint 1

The bag is a mystery box. Before you draw, you do not know the composition. You must define a set of hypotheses

H_{i}

, where

i

represents the number of white balls. What are the logical boundaries for

i

given that you successfully drew 2 white and 2 black balls?

Hint 2

You are looking for the probability of a specific hypothesis given an observed event. This is the classic domain of Bayes' Theorem. Remember, the prior probability

P (H_{i})

is uniform, but the likelihood

P (E ∣ H_{i})

depends heavily on the composition of the bag.

Hint 3

Start by setting up the Bayes' equation:

P (H_{4} ∣ E) = \frac{P ( E ∣ H _{4} ) P ( H _{4} )}{\sum _{i = 2}^{6} P ( E ∣ H _{i} ) P ( H _{i} )}

. Since

P (H_{i})

is constant for all valid

i

, it cancels out. Your task is to calculate the likelihoods

P (E ∣ H_{i}) = \frac{( 2 i ) ( 2 8 - i )}{( 4 8 )}

for

i \in {2, 3, 4, 5, 6}

Visualized Solution

Initial Setup

A bag contains $8$ balls.
Each ball is either White or Black.
The exact composition is initially unknown.

Defining Hypotheses $H_{i}$

Let $H_{i}$ be the hypothesis that the bag contains $i$ White balls.
Consequently, it contains $(8 - i)$ Black balls.
$i$ can range from $0$ to $8$ .

The Observed Event $E$

Event $E$ : $4$ balls are drawn at random.
Result: $2$ White and $2$ Black balls are found.

Restricting Possible Compositions

Since we drew $2$ White balls, the bag must have at least $2$ White balls ( $i \geq 2$ ).
Since we drew $2$ Black balls, the bag must have at least $2$ Black balls ( $8 - i \geq 2 ⟹ i \leq 6$ ).
Therefore, possible values for $i$ are ${2, 3, 4, 5, 6}$ .

Prior Probabilities $P (H_{i})$

There are $5$ valid hypotheses: $H_{2}, H_{3}, H_{4}, H_{5}, H_{6}$ .
Since no other information is given, we assume they are equally likely.
$P (H_{i}) = \frac{1}{5}$ for each valid $i$ .

Applying Bayes' Theorem

We need the probability that the bag has equal White and Black balls.
This means $i = 4$ , so we want $P (H_{4} ∣ E)$ .
Using Bayes' Theorem:
$P (H_{4} ∣ E) = \frac{P ( H _{4} ) P ( E ∣ H _{4} )}{\sum _{i = 2}^{6} P ( H _{i} ) P ( E ∣ H _{i} )}$

Formula for $P (E ∣ H_{i})$

$P (E ∣ H_{i})$ is the probability of drawing $2$ White, $2$ Black from a bag with $i$ White and $(8 - i)$ Black balls.
$P (E ∣ H_{i}) = \frac{( 2 i ) ( 2 8 - i )}{( 4 8 )}$

Calculating Total Outcomes

Total ways to draw $4$ balls from $8$ :
$(4 8) = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}$
$(4 8) = 70$

Calculating $P (E ∣ H_{2})$ and $P (E ∣ H_{6})$

For $i = 2$ : $P (E ∣ H_{2}) = \frac{( 2 2 ) ( 2 6 )}{70} = \frac{1 \times 15}{70} = \frac{15}{70}$
For $i = 6$ : $P (E ∣ H_{6}) = \frac{( 2 6 ) ( 2 2 )}{70} = \frac{15 \times 1}{70} = \frac{15}{70}$

Calculating $P (E ∣ H_{3})$ and $P (E ∣ H_{5})$

For $i = 3$ : $P (E ∣ H_{3}) = \frac{( 2 3 ) ( 2 5 )}{70} = \frac{3 \times 10}{70} = \frac{30}{70}$
For $i = 5$ : $P (E ∣ H_{5}) = \frac{( 2 5 ) ( 2 3 )}{70} = \frac{10 \times 3}{70} = \frac{30}{70}$

Calculating $P (E ∣ H_{4})$

For $i = 4$ : $P (E ∣ H_{4}) = \frac{( 2 4 ) ( 2 4 )}{70}$
$(2 4) = 6$
$P (E ∣ H_{4}) = \frac{6 \times 6}{70} = \frac{36}{70}$

Substituting Values

Since $P (H_{i}) = \frac{1}{5}$ for all $i$ , it cancels out from the numerator and denominator.
$P (H_{4} ∣ E) = \frac{P ( E ∣ H _{4} )}{P ( E ∣ H _{2} ) + P ( E ∣ H _{3} ) + P ( E ∣ H _{4} ) + P ( E ∣ H _{5} ) + P ( E ∣ H _{6} )}$
$P (H_{4} ∣ E) = \frac{\frac{36}{70}}{\frac{15}{70} + \frac{30}{70} + \frac{36}{70} + \frac{30}{70} + \frac{15}{70}}$

Simplifying the Expression

The common denominator $70$ also cancels out.
Numerator: $36$
Denominator: $15 + 30 + 36 + 30 + 15 = 126$
$P (H_{4} ∣ E) = \frac{36}{126}$

Final Answer

Simplify the fraction $\frac{36}{126}$ .
Both numbers are divisible by $18$ .
$\frac{36 \div 18}{126 \div 18} = \frac{2}{7}$
The probability that the bag contains equal number of white and black balls is $\frac{2}{7}$ .

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The Sigma Insight: Bayes' Theorem

Solution Diagram

The Mystery of the Bag

A Journey into Bayes' Theorem

Imagine you are standing in a room with a mysterious bag containing exactly 8 balls. You know they are either white or black, but you have no idea about the ratio.

In the world of JEE Advanced, we often encounter problems that feel like detective work. We are given an observation, and we must work backward to determine the most likely cause. This is the essence of Bayes' Theorem.

Defining the Universe of Possibilities

Let us define our hypotheses. Let

H_{i}

be the hypothesis that the bag contains exactly

i

white balls. Since there are 8 balls in total, the number of black balls is fixed at

8 - i

We have an observation: we drew 4 balls and found exactly 2 white and 2 black. Let this event be

E

. This event acts as a filter on our sample space.

If we drew 2 white balls, the bag must have contained at least 2 white balls (

i \geq 2

). If we drew 2 black balls, the bag must have contained at least 2 black balls, meaning

8 - i \geq 2

, or

i \leq 6

. Our universe of possibilities is thus

i \in {2, 3, 4, 5, 6}

The Bayes' Engine

We want to find the probability that the bag contains an equal number of white and black balls, which corresponds to

i = 4

. We are looking for

P (H_{4} ∣ E)

Bayes' Theorem tells us:

P (H_{4} ∣ E) = \frac{P ( E ∣ H _{4} ) P ( H _{4} )}{\sum _{i = 2}^{6} P ( E ∣ H _{i} ) P ( H _{i} )}

Since we have no prior information, we assume each valid hypothesis is equally likely, so

P (H_{i}) = \frac{1}{5}

. Because this term appears in every part of the numerator and denominator, it cancels out. We are left with the ratio of the likelihoods.

Calculating the Likelihoods

The probability of drawing 2 white and 2 black balls from a bag with

i

white and

8 - i

black balls is given by the hypergeometric distribution:

P (E ∣ H_{i}) = \frac{( 2 i ) ( 2 8 - i )}{( 4 8 )}

The denominator,

(4 8)

, is the total number of ways to choose 4 balls from 8, which is

70

. Now, we calculate the numerators for each case:

For $i = 2$ and $i = 6$ : $(2 2) (2 6) = 1 \times 15 = 15$ . Thus, $P (E ∣ H_{2}) = P (E ∣ H_{6}) = \frac{15}{70}$ . For

i = 3

and

i = 5

(2 3) (2 5) = 3 \times 10 = 30

. Thus,

P (E ∣ H_{3}) = P (E ∣ H_{5}) = \frac{30}{70}

. * For

i = 4

(2 4) (2 4) = 6 \times 6 = 36

. Thus,

P (E ∣ H_{4}) = \frac{36}{70}

The Final Synthesis

Now, we plug these values into our simplified Bayes' formula. The denominator is the sum of all these probabilities:

15 + 30 + 36 + 30 + 15 = 126

The numerator is our target case, which is

36

. Thus, the probability is:

P (H_{4} ∣ E) = \frac{36}{126}

Simplifying this fraction by dividing both the numerator and denominator by 18, we arrive at the elegant result:

$\frac{2}{7}$

This problem teaches us that in probability, we must always respect the constraints imposed by our observations. By systematically defining our hypotheses and applying the power of Bayes' Theorem, we turned a complex mystery into a clear, logical path.

Common Traps & FAQs

Why do we assume

P (H_{i}) = \frac{1}{5}

? Shouldn't it be

\frac{1}{9}

since there are 9 possibilities (0 to 8)?

Great question! While there are 9 possible compositions initially, the observed event (drawing 2 white and 2 black) acts as a filter. We are calculating the probability *given* that the event occurred. The hypotheses

H_{0}, H_{1}, H_{7}, H_{8}

are impossible because you cannot draw 2 white and 2 black balls from them. Thus, we only consider the valid subset

{H_{2}, H_{3}, H_{4}, H_{5}, H_{6}}

Why does the prior probability

P (H_{i})

cancel out?

Because we have no information about the bag's composition *before* the draw, we treat each valid hypothesis as equally likely (Principle of Indifference). Since

P (H_{i}) = k

for all

i

, the numerator becomes

k \cdot P (E ∣ H_{4})

and the denominator becomes

\sum k \cdot P (E ∣ H_{i})

. The constant

k

factors out and cancels, leaving us with the ratio of the likelihoods.

Is there a shortcut to calculating all those combinations?

Yes! Notice the symmetry. The probability of drawing 2 white and 2 black balls from a bag with

i

white balls is identical to drawing them from a bag with

(8 - i)

white balls. This means

P (E ∣ H_{2}) = P (E ∣ H_{6})

and

P (E ∣ H_{3}) = P (E ∣ H_{5})

. Use this symmetry to save time and reduce calculation errors!

Select Answer:

Visualized Solution

Initial Setup

Defining Hypotheses $H_{i}$

The Observed Event $E$

Restricting Possible Compositions

Prior Probabilities $P (H_{i})$

Applying Bayes' Theorem

Formula for $P (E ∣ H_{i})$

Calculating Total Outcomes

Calculating $P (E ∣ H_{2})$ and $P (E ∣ H_{6})$

Calculating $P (E ∣ H_{3})$ and $P (E ∣ H_{5})$

Calculating $P (E ∣ H_{4})$

Substituting Values

Simplifying the Expression

Final Answer

The Sigma Insight: Bayes' Theorem

The Mystery of the Bag

Defining the Universe of Possibilities

The Bayes' Engine

Calculating the Likelihoods

The Final Synthesis

Similar Questions

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

Bag $B_{1}$ contains 6 white and 4 blue balls, Bag $B_{2}$ contains 4 white and 6 blue balls, and Bag $B_{3}$ contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability, that the ball is drawn from Bag $B_{2}$ is:

Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and $n$ white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is $\frac{6}{11}$ , then $n$ is equal to ____.

A bag contains 10 balls out of which $k$ are red and $(10 - k)$ are black, where $0 \leq k \leq 10$ . If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

Comprehension Passage

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is

If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box $B_{2}$ is

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

Select Answer:

Visualized Solution

Initial Setup

Defining Hypotheses .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Hi​

The Observed Event .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E

Restricting Possible Compositions

Prior Probabilities .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(Hi​)

Applying Bayes' Theorem

Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣Hi​)

Calculating Total Outcomes

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣H2​) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣H6​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣H3​) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣H5​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣H4​)

Substituting Values

Simplifying the Expression

Final Answer

The Sigma Insight: Bayes' Theorem

The Mystery of the Bag

Defining the Universe of Possibilities

The Bayes' Engine

Calculating the Likelihoods

The Final Synthesis

Similar Questions

Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1 ball is drawn from each of the boxes B1​,B2​ and B3​, the probability that all 3 drawn balls are of the same colour is

Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is:

Defining Hypotheses $H_{i}$

The Observed Event $E$

Prior Probabilities $P (H_{i})$

Formula for $P (E ∣ H_{i})$

Calculating $P (E ∣ H_{2})$ and $P (E ∣ H_{6})$

Calculating $P (E ∣ H_{3})$ and $P (E ∣ H_{5})$

Calculating $P (E ∣ H_{4})$

If 1 ball is drawn from each of the boxes $B_{1}, B_{2}$ and $B_{3}$ , the probability that all 3 drawn balls are of the same colour is