Let U1 and U2 be two urns such that U1...

MathematicsBayes' TheoremJEE Advanced 2011Moderate

Comprehension Passage

Let

U_{1}

and

U_{2}

be two urns such that

U_{1}

contains 3 white and 2 red balls, and

U_{2}

contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from

U_{1}

and put into

U_{2}

. However, if tail appears then 2 balls are drawn at random from

U_{1}

and put into

U_{2}

. Now 1 ball is drawn at random from

U_{2}

Question 1:

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

(A)

13/30

(B)

23/30

(C)

19/30

(D)

11/30

Answer:B

Question 2:

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

(A)

17/23

(B)

11/23

(C)

15/23

(D)

12/23

Answer:D

Visualized Solution (Hindi)

Understanding the Setup

Urn $U_{1}$ : $3$ White ( $W$ ), $2$ Red ( $R$ ) balls.
Urn $U_{2}$ : $1$ White ( $W$ ) ball.
Coin Toss Outcomes: Head ( $H$ ) or Tail ( $T$ ).

Defining Event Probabilities

Probability of Head: $P (H) = \frac{1}{2}$
Probability of Tail: $P (T) = \frac{1}{2}$
Case 1 (Head): Transfer $1$ ball from $U_{1}$ to $U_{2}$ .
Case 2 (Tail): Transfer $2$ balls from $U_{1}$ to $U_{2}$ .

Case 1: Head Appeared (.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H)

If $H$ occurs, $1$ ball is transferred from $U_{1}$ ( $3 W, 2 R$ ).
Probability of transferring $W$ : $\frac{3}{5}$ . $U_{2}$ becomes ( $2 W, 0 R$ ).
Probability of transferring $R$ : $\frac{2}{5}$ . $U_{2}$ becomes ( $1 W, 1 R$ ).

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

$P (W ∣ H) = P (W_{t r an s}) \cdot P (W_{f r o m U 2} ∣ W_{t r an s}) + P (R_{t r an s}) \cdot P (W_{f r o m U 2} ∣ R_{t r an s})$
$P (W ∣ H) = (\frac{3}{5} \cdot \frac{2}{2}) + (\frac{2}{5} \cdot \frac{1}{2})$
$P (W ∣ H) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$

Case 2: Tail Appeared (.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T)

If $T$ occurs, $2$ balls are transferred from $U_{1}$ ( $3 W, 2 R$ ).
Prob( $2 W$ ): $\frac{^{3} C _{2}}{^{5} C _{2}} = \frac{3}{10}$ . $U_{2}$ becomes ( $3 W, 0 R$ ).
Prob( $2 R$ ): $\frac{^{2} C _{2}}{^{5} C _{2}} = \frac{1}{10}$ . $U_{2}$ becomes ( $1 W, 2 R$ ).
Prob( $1 W, 1 R$ ): $\frac{^{3} C _{1} \cdot ^{2} C _{1}}{^{5} C _{2}} = \frac{6}{10}$ . $U_{2}$ becomes ( $2 W, 1 R$ ).

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

$P (W ∣ T) = (\frac{3}{10} \cdot 1) + (\frac{1}{10} \cdot \frac{1}{3}) + (\frac{6}{10} \cdot \frac{2}{3})$
$P (W ∣ T) = \frac{3}{10} + \frac{1}{30} + \frac{12}{30}$
$P (W ∣ T) = \frac{9 + 1 + 12}{30} = \frac{22}{30} = \frac{11}{15}$

Total Probability of White Ball

Using Law of Total Probability:
$P (W) = P (H) P (W ∣ H) + P (T) P (W ∣ T)$
$P (W) = (\frac{1}{2} \cdot \frac{4}{5}) + (\frac{1}{2} \cdot \frac{11}{15})$
$P (W) = \frac{2}{5} + \frac{11}{30} = \frac{12 + 11}{30} = \frac{23}{30}$

Applying Bayes' Theorem

We need to find $P (H ∣ W)$ .
By Bayes' Theorem:
$P (H ∣ W) = \frac{P ( H ) \cdot P ( W ∣ H )}{P ( W )}$

Final Calculation

$P (H ∣ W) = \frac{\frac{1}{2} \cdot \frac{4}{5}}{23/30}$
$P (H ∣ W) = \frac{2/5}{23/30} = \frac{2}{5} \cdot \frac{30}{23}$
$P (H ∣ W) = \frac{12}{23}$

Key Takeaways

Total Probability $P (W) = \frac{23}{30}$ accounts for all possible paths.
Bayes' Theorem $P (H ∣ W) = \frac{12}{23}$ finds the probability of a 'cause' given an 'effect'.
JEE Tip: Always break down the problem into mutually exclusive cases (Head vs Tail) first.

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Conceptually Similar Problems

MathematicsBayes' TheoremJEE Advanced 2006Moderate

View in:English Hindi

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

MathematicsTotal Probability TheoremJEE Advanced 2001Easy

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An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. What is the probability that the ball drawn now is white?

Answer:

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MathematicsTotal Probability TheoremJEE Advanced 1988Moderate

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Urn $A$ contains 6 red and 4 black balls and urn $B$ contains 4 red and 6 black balls. One ball is drawn at random from urn $A$ and placed in urn $B$ . Then one ball is drawn at random from urn $B$ and placed in urn $A$ . If one ball is now drawn at random from urn $A$ , the probability that it is found to be red is .........

Answer:32/55

MathematicsClassical Definition of ProbabilityJEE Main 2010Easy

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An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is

(A)

2/7

(B)

1/21

(C)

2/23

(D)

1/3

Answer:A

MathematicsBayes' TheoremJEE Advanced 2015Moderate

View in:English Hindi

Comprehension Passage

Let

n_{1}

and

n_{2}

be the number of red and black balls, respectively, in box I. Let

n_{3}

and

n_{4}

be the number of red and black balls, respectively, in box II.

Question 1:

One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 1/3, then the correct option(s) with the possible values of $n_{1}, n_{2}, n_{3}$ and $n_{4}$ is(are)

(A)

n_{1} = 3, n_{2} = 3, n_{3} = 5, n_{4} = 15

(B)

n_{1} = 3, n_{2} = 6, n_{3} = 10, n_{4} = 50

(C)

n_{1} = 8, n_{2} = 6, n_{3} = 5, n_{4} = 20

(D)

n_{1} = 6, n_{2} = 12, n_{3} = 5, n_{4} = 20

Answer:A, B

Question 2:

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)

(A)

n_{1} = 4

and

n_{2} = 6

(B)

n_{1} = 2

and

n_{2} = 3

(C)

n_{1} = 10

and

n_{2} = 20

(D)

n_{1} = 3

and

n_{2} = 6

Answer:C, D

MathematicsBayes' TheoremJEE Advanced 2013Moderate

View in:English Hindi

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

(A)

82/648

(B)

90/648

(C)

558/648

(D)

566/648

Answer:A

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

(A)

116/181

(B)

126/181

(C)

65/181

(D)

55/181

Answer:D

MathematicsBayes' TheoremJEE Advanced 2002Moderate

View in:English Hindi

A box contains $N$ coins, $m$ of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is $1/2$ , while it is $2/3$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair?

Answer:

\frac{9 m}{m + 8 N}

MathematicsAddition and Multiplication TheoremsJEE Advanced 1998Easy

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If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is

(A)

13/32

(B)

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(C)

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(D)

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MathematicsCombinations and SelectionJEE Main 2010Easy

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There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

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(B)

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A box contains 12 red and 6 white balls. Balls are drawn from the box one at a time without replacement. If in 6 draws there are at least 4 white balls, find the probability that exactly one white is drawn in the next two draws. (binomial coefficients can be left as such)

Answer:

\frac{^{10} C _{1} \times ^{2} C _{1}}{^{12} C _{2}} \times \frac{^{12} C _{4} \times ^{6} C _{2}}{^{18} C _{6}} + \frac{^{11} C _{1} \times ^{1} C _{1}}{^{12} C _{2}} \times \frac{^{12} C _{5} \times ^{6} C _{1}}{^{18} C _{6}}

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

Visualized Solution (Hindi)

Understanding the Setup

Defining Event Probabilities

Case 1: Head Appeared ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H)

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

Case 2: Tail Appeared ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T)

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

Total Probability of White Ball

Applying Bayes' Theorem

Final Calculation

Key Takeaways

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.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

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Calculating $P (W ∣ H)$

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