JEE Main 2025 April

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Animated Solution for Mathematics - Probability: Given three identical bags each containing 10 balls, whose colours are as follows : \begin{array}{|l|l|l|l|} \hline & \textbf{Red} & \textbf{Blue} & \textbf{Green} \\ \hline Bag I & 3 & 2 & 5 \\ \hline Bag II & 4 & 3 & 3 \\ \hline Bag III & 5 & 1 & 4 \\ \hline \end{array} A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$ , then the value of $(\frac{1}{p} + \frac{1}{q})$ is :

Select Answer:

Progressive Hints

Hint 1

Think of this as a detective story. You are given the 'crime scene' (the color of the ball) and you need to trace it back to the 'suspect' (the bag). This is the essence of Bayes' Theorem: reverse probability. Visualize the tree diagram where the first branch is the choice of the bag, and the second branch is the color of the ball.

Hint 2

Recall that the probability of choosing any bag is equal,

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

. When applying Bayes' Theorem, notice how this prior probability appears in every term of the numerator and denominator. This is a massive hint that the math is about to get much simpler.

Hint 3

To find

p = P (B_{I} ∣ R)

, use the formula

p = \frac{P ( B _{I} ) \cdot P ( R ∣ B _{I} )}{P ( B _{I} ) P ( R ∣ B _{I} ) + P ( B _{I I} ) P ( R ∣ B _{I I} ) + P ( B _{I I I} ) P ( R ∣ B _{I I I} )}

. Substitute the counts directly, and watch how the common factors simplify your life. Do the same for

q = P (B_{I I I} ∣ G)

Visualized Solution

Visualizing the Setup

Three identical bags: $B_{I}, B_{I I}, B_{I I I}$
Each bag contains $10$ balls.
Bag I: $3 R, 2 B, 5 G$
Bag II: $4 R, 3 B, 3 G$
Bag III: $5 R, 1 B, 4 G$

Probability of Choosing a Bag

Since bags are chosen at random:
$P (B_{I}) = P (B_{I I}) = P (B_{I I I}) = \frac{1}{3}$

Defining Probability $p$

$p = P (B_{I} ∣ R)$
By Bayes' Theorem:
$p = \frac{P ( B _{I} ) \cdot P ( R ∣ B _{I} )}{P ( B _{I} ) P ( R ∣ B _{I} ) + P ( B _{I I} ) P ( R ∣ B _{I I} ) + P ( B _{I I I} ) P ( R ∣ B _{I I I} )}$

Conditional Probabilities for Red Balls

$P (R ∣ B_{I}) = \frac{3}{10}$
$P (R ∣ B_{I I}) = \frac{4}{10}$
$P (R ∣ B_{I I I}) = \frac{5}{10}$

Calculating $p$ (Substitution)

$p = \frac{\frac{1}{3} \cdot \frac{3}{10}}{\frac{1}{3} \cdot \frac{3}{10} + \frac{1}{3} \cdot \frac{4}{10} + \frac{1}{3} \cdot \frac{5}{10}}$

Calculating $p$ (Simplification)

Cancel common terms $\frac{1}{3}$ and $\frac{1}{10}$ :
$p = \frac{3}{3 + 4 + 5}$

Final Value of $p$

$p = \frac{3}{12}$
$p = \frac{1}{4}$

Defining Probability $q$

$q = P (B_{I I I} ∣ G)$
By Bayes' Theorem:
$q = \frac{P ( B _{I I I} ) \cdot P ( G ∣ B _{I I I} )}{P ( B _{I} ) P ( G ∣ B _{I} ) + P ( B _{I I} ) P ( G ∣ B _{I I} ) + P ( B _{I I I} ) P ( G ∣ B _{I I I} )}$

Conditional Probabilities for Green Balls

$P (G ∣ B_{I}) = \frac{5}{10}$
$P (G ∣ B_{I I}) = \frac{3}{10}$
$P (G ∣ B_{I I I}) = \frac{4}{10}$

Calculating $q$ (Substitution)

$q = \frac{\frac{1}{3} \cdot \frac{4}{10}}{\frac{1}{3} \cdot \frac{5}{10} + \frac{1}{3} \cdot \frac{3}{10} + \frac{1}{3} \cdot \frac{4}{10}}$

Calculating $q$ (Simplification)

Cancel common terms $\frac{1}{3}$ and $\frac{1}{10}$ :
$q = \frac{4}{5 + 3 + 4}$

Final Value of $q$

$q = \frac{4}{12}$
$q = \frac{1}{3}$

Final Calculation

We have $p = \frac{1}{4}$ and $q = \frac{1}{3}$
$\frac{1}{p} = 4$
$\frac{1}{q} = 3$
$\frac{1}{p} + \frac{1}{q} = 4 + 3 = 7$

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The Sigma Insight: Bayes' Theorem

Solution Diagram

Analyzing the Setup

We are tasked with identifying the source of a ball drawn from one of three identical bags,

B_{I}

B_{I I}

, and

B_{I I I}

. Since the bags are chosen at random, the prior probability for selecting any bag is equal:

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

Calculating the Probability of a Red Ball

We define

p = P (B_{I} ∣ R)

as the probability that the ball came from Bag I, given that it is red. According to Bayes' Theorem:

p = \frac{P ( B _{I} ) \cdot P ( R ∣ B _{I} )}{P ( B _{I} ) P ( R ∣ B _{I} ) + P ( B _{I I} ) P ( R ∣ B _{I I} ) + P ( B _{I I I} ) P ( R ∣ B _{I I I} )}

The conditional probabilities for drawing a red ball from each bag are

P (R ∣ B_{I}) = \frac{3}{10}

P (R ∣ B_{I I}) = \frac{4}{10}

, and

P (R ∣ B_{I I I}) = \frac{5}{10}

Substituting these values, the common factors of

\frac{1}{3}

and

\frac{1}{10}

cancel out across the numerator and denominator:

p = \frac{3}{3 + 4 + 5} = \frac{3}{12} = \frac{1}{4}

Calculating the Probability of a Green Ball

Next, we define

q = P (B_{I I I} ∣ G)

as the probability that the ball came from Bag III, given that it is green. The conditional probabilities for drawing a green ball are

P (G ∣ B_{I}) = \frac{5}{10}

P (G ∣ B_{I I}) = \frac{3}{10}

, and

P (G ∣ B_{I I I}) = \frac{4}{10}

Applying Bayes' Theorem again:

q = \frac{P ( B _{I I I} ) \cdot P ( G ∣ B _{I I I} )}{P ( B _{I} ) P ( G ∣ B _{I} ) + P ( B _{I I} ) P ( G ∣ B _{I I} ) + P ( B _{I I I} ) P ( G ∣ B _{I I I} )}

Following the same simplification process where the common factors cancel, we obtain:

q = \frac{4}{5 + 3 + 4} = \frac{4}{12} = \frac{1}{3}

Final Calculation

With our values determined as

p = \frac{1}{4}

and

q = \frac{1}{3}

, we perform the final arithmetic operation requested:

\frac{1}{p} + \frac{1}{q} = 4 + 3 = 7

The final result is 7.

Common Traps & FAQs

Wait, why do we cancel the

\frac{1}{3}

? Is that always allowed?

Great question! Yes, because the bags are chosen at random, the prior probability is uniform. Since

\frac{1}{3}

is a common factor in every term of the numerator and the denominator, it cancels out perfectly. It is a beautiful simplification that happens when the 'prior' is unbiased. You are essentially comparing the number of favorable outcomes directly.

Can I just use the number of red balls instead of the full probability formula?

You are thinking like a pro! Yes, because the total number of balls in each bag is the same (10) and the probability of picking each bag is the same (1/3), the ratio of probabilities is simply the ratio of the number of favorable balls. It is a shortcut, but always write out the Bayes' formula first to be safe and to ensure you don't miss any hidden variables!

What if the bags had different total numbers of balls? Would the answer change?

Absolutely. If the bags had different totals, the conditional probabilities

P (R ∣ B_{i})

would change, and the

1/10

factor wouldn't cancel out as easily. Always check the total count in each bag before assuming you can just use the raw counts. Never let the pattern blind you to the underlying data.

Select Answer:

Visualized Solution

Visualizing the Setup

Probability of Choosing a Bag

Defining Probability $p$

Conditional Probabilities for Red Balls

Calculating $p$ (Substitution)

Calculating $p$ (Simplification)

Final Value of $p$

Defining Probability $q$

Conditional Probabilities for Green Balls

Calculating $q$ (Substitution)

Calculating $q$ (Simplification)

Final Value of $q$

Final Calculation

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Calculating the Probability of a Red Ball

Calculating the Probability of a Green Ball

Final Calculation

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 :

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 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:

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

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 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

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)

Select Answer:

Visualized Solution

Visualizing the Setup

Probability of Choosing a Bag

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

Conditional Probabilities for Red Balls

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

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

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

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

Conditional Probabilities for Green Balls

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

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

Final Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Final Calculation

The Sigma Insight: Bayes' Theorem

Analyzing the Setup

Calculating the Probability of a Red Ball

Calculating the Probability of a Green Ball

Final Calculation

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 :

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 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

Defining Probability $p$

Calculating $p$ (Substitution)

Calculating $p$ (Simplification)

Final Value of $p$

Defining Probability $q$

Calculating $q$ (Substitution)

Calculating $q$ (Simplification)

Final Value of $q$

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