JEE Main 2024 (04 Apr Shift 1)

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Animated Solution for Mathematics - Probability: Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Q: What if the urns had different total numbers of balls?

If the urns had different totals, the $1/12$ would not be a common factor. You would have to calculate the total probability of drawing a black ball by summing the individual probabilities $P(E_i)P(B|E_i)$ without canceling anything. Always be careful to check the total number of balls in each case!

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. You are dealing with an inverse probability problem. You know the final outcome (a black ball was drawn), but you need to determine the source (which urn it came from). This is the classic setup for Bayes' Theorem, where we update our prior beliefs based on new evidence.

Hint 2

Hint 2: The Arsenal. Recall the formula for Bayes' Theorem:

P (E_{i} ∣ B) = \frac{P ( E _{i} ) P ( B ∣ E _{i} )}{\sum _{j = 1}^{3} P ( E _{j} ) P ( B ∣ E _{j} )}

. Here,

E_{1}, E_{2}, E_{3}

represent selecting Urn A, B, and C, and

B

is the event of drawing a black ball. Since each urn is selected at random,

P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}

Hint 3

Hint 3: The Execution. Start by calculating the conditional probability of drawing a black ball from each urn:

P (B ∣ E_{1}) = \frac{5}{12}

P (B ∣ E_{2}) = \frac{7}{12}

, and

P (B ∣ E_{3}) = \frac{6}{12}

. Now, substitute these into the Bayes' formula. Notice how the common factor of

\frac{1}{3}

and the denominator of

12

will simplify the expression significantly.

Visualized Solution

Visualizing the Urns

Three urns: A, B, and C.
Each urn contains a mix of Red and Black balls.
Goal: Find $P (Urn A ∣ Black Ball)$ .
This is a classic application of Bayes' Theorem.

Defining Selection Events

Let $E_{1}, E_{2}, E_{3}$ be the events of selecting urns A, B, and C respectively.
Since selection is random: $P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}$ .

Defining the Condition

Let $B$ be the event that the ball drawn is black.
We need to find the conditional probability $P (E_{1} ∣ B)$ .

Probabilities from Urn A

Urn A: $7$ Red, $5$ Black (Total = $12$ ).
Conditional probability: $P (B ∣ E_{1}) = \frac{5}{12}$ .

Probabilities from Urn B

Urn B: $5$ Red, $7$ Black (Total = $12$ ).
Conditional probability: $P (B ∣ E_{2}) = \frac{7}{12}$ .

Probabilities from Urn C

Urn C: $6$ Red, $6$ Black (Total = $12$ ).
Conditional probability: $P (B ∣ E_{3}) = \frac{6}{12}$ .

Applying Bayes' Theorem

By Bayes' Theorem:
$P (E_{1} ∣ B) = \frac{P ( E _{1} ) P ( B ∣ E _{1} )}{P ( E _{1} ) P ( B ∣ E _{1} ) + P ( E _{2} ) P ( B ∣ E _{2} ) + P ( E _{3} ) P ( B ∣ E _{3} )}$

Raw Substitution

Substitute the values:
$P (E_{1} ∣ B) = \frac{\frac{1}{3} \cdot \frac{5}{12}}{\frac{1}{3} \cdot \frac{5}{12} + \frac{1}{3} \cdot \frac{7}{12} + \frac{1}{3} \cdot \frac{6}{12}}$

Simplifying the Expression

Cancel the common factor $\frac{1}{3} \cdot \frac{1}{12}$ from numerator and denominator:
$P (E_{1} ∣ B) = \frac{5}{5 + 7 + 6}$

Final Calculation

Sum the denominator: $5 + 7 + 6 = 18$ .
Final Result: $P (E_{1} ∣ B) = \frac{5}{18}$ .

Conclusion and Key Takeaway

Key Takeaway: Bayes' Theorem relates current evidence to prior probabilities.
Final Answer: $\frac{5}{18}$

00:00 / 00:00

The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

Imagine you are standing in a room with three mysterious urns, labeled

A

B

, and

C

. Each urn is a container of secrets, holding a specific mixture of red and black balls.

Urn

A

holds

7

red and

5

black balls. Urn

B

holds

5

red and

7

black balls. Urn

C

holds

6

red and

6

black balls.

You are asked to pick one urn at random and draw a single ball. The ball you draw is black. We must determine the probability that it came from Urn

A

using Bayes' Theorem.

Defining the Events

To solve this, we formalize our experiment. Let

E_{1}, E_{2}, E_{3}

be the events of selecting Urn

A

, Urn

B

, and Urn

C

, respectively.

Since the selection is random, each urn has an equal prior probability:

P (E_{1}) = P (E_{2}) = P (E_{3}) = \frac{1}{3}

Let

B

be the event that the ball drawn is black. We want to find the conditional probability

P (E_{1} ∣ B)

, which represents the probability of having chosen Urn

A

given that the ball is black.

The Conditional Probabilities

Next, we calculate the likelihood of drawing a black ball from each specific urn:

For Urn

A

P (B ∣ E_{1}) = \frac{5}{12}

For Urn

B

P (B ∣ E_{2}) = \frac{7}{12}

For Urn

C

P (B ∣ E_{3}) = \frac{6}{12}

Applying Bayes' Theorem

Bayes' Theorem provides the following formula to update our belief:

P (E_{1} ∣ B) = \frac{P ( E _{1} ) P ( B ∣ E _{1} )}{P ( E _{1} ) P ( B ∣ E _{1} ) + P ( E _{2} ) P ( B ∣ E _{2} ) + P ( E _{3} ) P ( B ∣ E _{3} )}

Substituting our known values into the equation, we obtain:

P (E_{1} ∣ B) = \frac{\frac{1}{3} \cdot \frac{5}{12}}{\frac{1}{3} \cdot \frac{5}{12} + \frac{1}{3} \cdot \frac{7}{12} + \frac{1}{3} \cdot \frac{6}{12}}

Final Calculation

Notice that the term

\frac{1}{3} \cdot \frac{1}{12}

is common to every part of the expression. We can factor it out and cancel it entirely from the numerator and the denominator.

This simplifies the expression to:

P (E_{1} ∣ B) = \frac{5}{5 + 7 + 6}

Calculating the sum in the denominator, we find

5 + 7 + 6 = 18

. Thus, the final probability is:

$P (E_{1} ∣ B) = \frac{5}{18}$

Common Traps & FAQs

Wait, why is the probability of picking each urn

1/3

The problem states that 'One of the urn is selected at random'. In probability, this phrase implies that each urn has an equal likelihood of being chosen. Since there are three urns, the probability for each is

1/3

Why can we just cancel out the

1/3

and the

1/12

That is a great observation! In the Bayes' formula, the term

P (E_{i}) P (B ∣ E_{i})

appears in both the numerator and every term of the denominator. Because

P (E_{i})

1/3

for all

i

, and the total number of balls in each urn is

12

, the term

\frac{1}{3} \cdot \frac{1}{12}

is a common factor. When you have a common factor in every term of a fraction, you can factor it out and cancel it, leaving you with a much simpler calculation.

What if the urns had different total numbers of balls?

If the urns had different totals, the

1/12

would not be a common factor. You would have to calculate the total probability of drawing a black ball by summing the individual probabilities

P (E_{i}) P (B ∣ E_{i})

without canceling anything. Always be careful to check the total number of balls in each case!

Animated Solution for Mathematics - Probability: Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Select Answer:

Visualized Solution

Visualizing the Urns

Defining Selection Events

Defining the Condition

Probabilities from Urn A

Probabilities from Urn B

Probabilities from Urn C

Applying Bayes' Theorem

Raw Substitution

Simplifying the Expression

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining the Events

The Conditional Probabilities

Applying Bayes' Theorem

Final Calculation

Similar Questions

Comprehension Passage

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

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 :

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:

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

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:

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:

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

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

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

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

Animated Solution for Mathematics - Probability: Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is :

Select Answer:

Visualized Solution

Visualizing the Urns

Defining Selection Events

Defining the Condition

Probabilities from Urn A

Probabilities from Urn B

Probabilities from Urn C

Applying Bayes' Theorem

Raw Substitution

Simplifying the Expression

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Defining the Events

The Conditional Probabilities

Applying Bayes' Theorem

Final Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The probability of the drawn ball from U2​ being white is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given that the drawn ball from U2​ is white, the probability that head appeared on the coin is

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 :

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:

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

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 P(ui​)∝i, where i=1,2,3,…,n, then limn→∞​P(w)=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(ui​)=c, (a constant) then P(un​/w)=

.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

The probability of the drawn ball from $U_{2}$ being white is

Given that the drawn ball from $U_{2}$ is white, the probability that head appeared on the coin is

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

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

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