JEE Advanced 2013

LEVELJEE Main

Animated Solution for Mathematics - Probability: Comprehension Passage

A box

B_{1}

contains 1 white ball, 3 red balls and 2 black balls. Another box

B_{2}

contains 2 white balls, 3 red balls and 4 black balls. A third box

B_{3}

contains 3 white balls, 4 red balls and 5 black balls.

Question 1:

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

Select Answer:

Question 2:

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

Select Answer:

Visualized Solution

Visualizing the Setup

Box $B_{1}$ : $1$ White, $3$ Red, $2$ Black (Total = $6$ balls)
Box $B_{2}$ : $2$ White, $3$ Red, $4$ Black (Total = $9$ balls)
Box $B_{3}$ : $3$ White, $4$ Red, $5$ Black (Total = $12$ balls)

Part 1: Probability of Same Color

We draw exactly one ball from each of the three boxes.
We want the probability that all three drawn balls have the same color.
This can happen in three mutually exclusive ways:
1. All three are White ( $W_{1} W_{2} W_{3}$ )
2. All three are Red ( $R_{1} R_{2} R_{3}$ )
3. All three are Black ( $B_{1} B_{2} B_{3}$ )

Case 1: Drawing Three White Balls

Probability of White from $B_{1}$ : $P (W_{1}) = \frac{1}{6}$
Probability of White from $B_{2}$ : $P (W_{2}) = \frac{2}{9}$
Probability of White from $B_{3}$ : $P (W_{3}) = \frac{3}{12}$
Since the draws are independent:
$P (W_{1} W_{2} W_{3}) = \frac{1}{6} \times \frac{2}{9} \times \frac{3}{12} = \frac{6}{648}$

Case 2: Drawing Three Red Balls

Probability of Red from $B_{1}$ : $P (R_{1}) = \frac{3}{6}$
Probability of Red from $B_{2}$ : $P (R_{2}) = \frac{3}{9}$
Probability of Red from $B_{3}$ : $P (R_{3}) = \frac{4}{12}$
Since the draws are independent:
$P (R_{1} R_{2} R_{3}) = \frac{3}{6} \times \frac{3}{9} \times \frac{4}{12} = \frac{36}{648}$

Case 3: Drawing Three Black Balls

Probability of Black from $B_{1}$ : $P (B_{1}) = \frac{2}{6}$
Probability of Black from $B_{2}$ : $P (B_{2}) = \frac{4}{9}$
Probability of Black from $B_{3}$ : $P (B_{3}) = \frac{5}{12}$
Since the draws are independent:
$P (B_{1} B_{2} B_{3}) = \frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} = \frac{40}{648}$

Total Probability of Same Color

Add the probabilities of the three mutually exclusive cases:
$P (Same Color) = \frac{6}{648} + \frac{36}{648} + \frac{40}{648}$
$P (Same Color) = \frac{82}{648}$
This matches Option A.

Part 2: Reverse Probability Setup

A box is selected at random: $P (B_{1}) = P (B_{2}) = P (B_{3}) = \frac{1}{3}$
Two balls are drawn without replacement from the selected box.
Event $E$ : One ball is White and the other is Red.
We want to find the posterior probability: $P (B_{2} ∣ E)$

Finding $P (E ∣ B_{1})$ and $P (E ∣ B_{2})$

For Box $B_{1}$ ( $1$ W, $3$ R, total $6$ ):
$P (E ∣ B_{1}) = \frac{( 1 1 ) \times ( 1 3 )}{( 2 6 )} = \frac{1 \times 3}{15} = \frac{3}{15} = \frac{1}{5}$
For Box $B_{2}$ ( $2$ W, $3$ R, total $9$ ):
$P (E ∣ B_{2}) = \frac{( 1 2 ) \times ( 1 3 )}{( 2 9 )} = \frac{2 \times 3}{36} = \frac{6}{36} = \frac{1}{6}$

Finding $P (E ∣ B_{3})$

For Box $B_{3}$ ( $3$ W, $4$ R, total $12$ ):
$P (E ∣ B_{3}) = \frac{( 1 3 ) \times ( 1 4 )}{( 2 12 )} = \frac{3 \times 4}{66} = \frac{12}{66} = \frac{2}{11}$

Applying Bayes' Formula

Substitute the values into Bayes' Formula:
$P (B_{2} ∣ E) = \frac{P ( E ∣ B _{2} ) P ( B _{2} )}{P ( E ∣ B _{1} ) P ( B _{1} ) + P ( E ∣ B _{2} ) P ( B _{2} ) + P ( E ∣ B _{3} ) P ( B _{3} )}$
Since $P (B_{i}) = \frac{1}{3}$ is common, it cancels out:
$P (B_{2} ∣ E) = \frac{\frac{1}{6}}{\frac{1}{5} + \frac{1}{6} + \frac{2}{11}}$

Final Calculation & Answer

Simplify the denominator:
$\frac{1}{5} + \frac{1}{6} + \frac{2}{11} = \frac{66 + 55 + 60}{330} = \frac{181}{330}$
Substitute back:
$P (B_{2} ∣ E) = \frac{\frac{1}{6}}{\frac{181}{330}} = \frac{1}{6} \times \frac{330}{181} = \frac{55}{181}$
This matches Option D.

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The Sigma Insight: Bayes' Theorem

Solution Diagram

The Probability Odyssey

Unlocking the Mystery of the Three Boxes

Welcome, fellow traveler, to the fascinating world of probability! Today, we are not just solving a problem; we are embarking on a journey through the logic of uncertainty. We have three boxes, each a treasure chest of colored balls, and we are going to use the power of mathematics to predict the future.

Phase 1

The Harmony of Colors

Imagine you are standing before three boxes:

B_{1}

B_{2}

, and

B_{3}

. Each box has its own unique personality, defined by its contents:

$B_{1}$ holds $1$ white, $3$ red, and $2$ black balls (total $6$ ).

B_{2}

holds

2

white,

3

red, and

4

black balls (total

9

). *

B_{3}

holds

3

white,

4

red, and

5

black balls (total

12

Our first challenge is to draw one ball from each box and ensure they all match in color. Because the draw from

B_{1}

has no impact on the draw from

B_{2}

B_{3}

, we treat these as independent events. We have three mutually exclusive paths to success: all white, all red, or all black.

For the white case, the probability is:

P (W_{1} \cap W_{2} \cap W_{3}) = \frac{1}{6} \times \frac{2}{9} \times \frac{3}{12} = \frac{6}{648}

For the red case, the probability is:

P (R_{1} \cap R_{2} \cap R_{3}) = \frac{3}{6} \times \frac{3}{9} \times \frac{4}{12} = \frac{36}{648}

For the black case, the probability is:

P (B_{1} \cap B_{2} \cap B_{3}) = \frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} = \frac{40}{648}

By summing these, we arrive at the total probability:

\frac{6 + 36 + 40}{648} = \frac{82}{648} = \frac{41}{324}

Phase 2

The Bayes' Theorem Twist

Now, the plot thickens. We shift from 'forward' probability to 'reverse' probability. We pick a box at random, draw two balls without replacement, and find one white and one red. We want to know: Which box did this come from?

We define event

E

as drawing one white and one red ball. We calculate the conditional probability

P (E ∣ B_{i})

for each box using combinations:

For

B_{1}

P (E ∣ B_{1}) = \frac{( 1 1 ) \times ( 1 3 )}{( 2 6 )} = \frac{3}{15} = \frac{1}{5}

For

B_{2}

P (E ∣ B_{2}) = \frac{( 1 2 ) \times ( 1 3 )}{( 2 9 )} = \frac{6}{36} = \frac{1}{6}

For

B_{3}

P (E ∣ B_{3}) = \frac{( 1 3 ) \times ( 1 4 )}{( 2 12 )} = \frac{12}{66} = \frac{2}{11}

The Grand Finale

The Power of Cancellation

Applying Bayes' Formula, we seek

P (B_{2} ∣ E)

. Since the prior probability of choosing any box is

P (B_{i}) = \frac{1}{3}

, it appears in every term of the formula and cancels out entirely.

We are left with the following expression:

P (B_{2} ∣ E) = \frac{\frac{1}{6}}{\frac{1}{5} + \frac{1}{6} + \frac{2}{11}}

Calculating the denominator:

\frac{1}{5} + \frac{1}{6} + \frac{2}{11} = \frac{66 + 55 + 60}{330} = \frac{181}{330}

Finally, we divide the numerator by this result:

P (B_{2} ∣ E) = \frac{1}{6} \times \frac{330}{181} = \frac{55}{181}

Through careful counting and the logical structure of Bayes' Theorem, we have unraveled the mystery. The final probability is $\frac{55}{181}$ .

Common Traps & FAQs

Wait, why do we add the probabilities in Part 1 instead of multiplying them?

Great question! We multiply when events must happen *together* (like drawing a white ball from Box 1 AND Box 2 AND Box 3). We add when we have *mutually exclusive cases* that satisfy the condition. Since you can't draw three white balls AND three red balls at the same time, these are separate scenarios. We calculate the probability of each scenario and add them to get the total probability.

Why can we just cancel out the

1/3

in Bayes' Theorem?

Bayes' Theorem involves the prior probability of selecting each box, which is

P (B_{1}) = P (B_{2}) = P (B_{3}) = 1/3

. Since this factor appears in every single term of the numerator and the denominator, it acts as a common multiplier. Mathematically, you can factor it out and divide it away, simplifying your calculation significantly!

Why use combinations

(r n)

for the second part?

We use combinations because the order in which we draw the two balls doesn't matter—we just need one white and one red. Since we are drawing *without replacement*, the total number of ways to pick 2 balls from a box with

N

balls is

(2 N)

. This is the most robust way to count the favorable outcomes versus the total sample space.

Animated Solution for Mathematics - Probability: 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

Select Answer:

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

Select Answer:

Visualized Solution

Visualizing the Setup

Part 1: Probability of Same Color

Case 1: Drawing Three White Balls

Case 2: Drawing Three Red Balls

Case 3: Drawing Three Black Balls

Total Probability of Same Color

Part 2: Reverse Probability Setup

Finding $P (E ∣ B_{1})$ and $P (E ∣ B_{2})$

Finding $P (E ∣ B_{3})$

Applying Bayes' Formula

Final Calculation & Answer

The Sigma Insight: Bayes' Theorem

The Probability Odyssey

Phase 1

Phase 2

The Grand Finale

Similar Questions

Comprehension Passage

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)

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 $B_{1}$ 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 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:

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:

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 ____.

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

Animated Solution for Mathematics - Probability: 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

Select Answer:

Select Answer:

Visualized Solution

Visualizing the Setup

Part 1: Probability of Same Color

Case 1: Drawing Three White Balls

Case 2: Drawing Three Red Balls

Case 3: Drawing Three Black Balls

Total Probability of Same Color

Part 2: Reverse Probability Setup

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣B1​) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣B2​)

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(E∣B3​)

Applying Bayes' Formula

Final Calculation & Answer

The Sigma Insight: Bayes' Theorem

The Probability Odyssey

Phase 1

Phase 2

The Grand Finale

Similar Questions

Comprehension Passage

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:

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:

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

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

Finding $P (E ∣ B_{1})$ and $P (E ∣ B_{2})$

Finding $P (E ∣ B_{3})$

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