A box B1 contains 1 white ball, 3 red...

MathematicsBayes' TheoremJEE Advanced 2013Moderate

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

Visualized Solution (English)

Visualizing the Boxes

Box $B_{1}$ : 1W, 3R, 2B (Total = 6)
Box $B_{2}$ : 2W, 3R, 4B (Total = 9)
Box $B_{3}$ : 3W, 4R, 5B (Total = 12)
Event $E$ : All 3 drawn balls have the same color.

Case 1: All White Balls

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

Case 2 & 3: Red and Black

$P (R_{1} R_{2} R_{3}) = \frac{3}{6} \times \frac{3}{9} \times \frac{4}{12} = \frac{36}{648}$
$P (B_{1} B_{2} B_{3}) = \frac{2}{6} \times \frac{4}{9} \times \frac{5}{12} = \frac{40}{648}$

Final Probability for Part 1

$P (Same Color) = \frac{6}{648} + \frac{36}{648} + \frac{40}{648}$
$P (Same Color) = \frac{82}{648}$

Transition to Bayes' Theorem

Let $E$ be the event: 1 White and 1 Red ball are drawn.
Let $B_{i}$ be the event of selecting Box $i$ .
$P (B_{1}) = P (B_{2}) = P (B_{3}) = \frac{1}{3}$

Conditional Probabilities .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (E ∣ B_{i})

$P (E ∣ B_{1}) = \frac{( 1 1 ) \times ( 1 3 )}{( 2 6 )} = \frac{3}{15} = \frac{1}{5}$
$P (E ∣ B_{2}) = \frac{( 1 2 ) \times ( 1 3 )}{( 2 9 )} = \frac{6}{36} = \frac{1}{6}$
$P (E ∣ B_{3}) = \frac{( 1 3 ) \times ( 1 4 )}{( 2 12 )} = \frac{12}{66} = \frac{2}{11}$

Applying Bayes' Formula

$P (B_{2} ∣ E) = \frac{P ( E ∣ B _{2} ) P ( B _{2} )}{\sum P ( E ∣ B _{i} ) P ( B _{i} )}$
$P (B_{2} ∣ E) = \frac{\frac{1}{6} \times \frac{1}{3}}{\frac{1}{3} ( \frac{1}{5} + \frac{1}{6} + \frac{2}{11} )}$

The Final Result

$P (B_{2} ∣ E) = \frac{1/6}{181/330} = \frac{1}{6} \times \frac{330}{181}$
Final Answer: $55/181$
Key Takeaway: Bayes' Theorem is essential for finding the cause (box) given the effect (drawn balls).

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Conceptually Similar Problems

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)

1/4

(C)

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(D)

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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

MathematicsTotal Probability TheoremJEE Advanced 2004Moderate

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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}}

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A box contains $24$ identical balls of which $12$ are white and $12$ are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the $4$ th time on the $7$ th draw is

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If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is

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220 (1/3)^{12}

(B)

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(C)

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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

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1/21

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2/23

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MathematicsBayes' TheoremJEE Advanced 2011Moderate

View in:English Hindi

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

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

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

Visualized Solution (English)

Visualizing the Boxes

Case 1: All White Balls

Case 2 & 3: Red and Black

Final Probability for Part 1

Transition to Bayes' Theorem

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

Applying Bayes' Formula

The Final Result

Conceptually Similar Problems

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

Comprehension Passage

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)

Balls are drawn one-by-one without replacement from a box containing 2 black, 4 white and 3 red balls till all the balls are drawn. Find the probability that the balls drawn are in the order 2 black, 4 white and 3 red.

A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

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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

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Comprehension Passage

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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

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