Box 1 contains three cards bearing...

MathematicsClassical Definition of ProbabilityJEE Advanced 2014Moderate

Comprehension Passage

Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5; and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes. Let

x_{i}

be number on the card drawn from the

i^{t h}

box,

i = 1, 2, 3

Question 1:

The probability that $x_{1} + x_{2} + x_{3}$ is odd, is

(A)

29/105

(B)

53/105

(C)

57/105

(D)

1/2

Answer:B

Question 2:

The probability that $x_{1}, x_{2}, x_{3}$ are in an arithmetic progression, is

(A)

9/105

(B)

10/105

(C)

11/105

(D)

7/105

Answer:C

Visualized Solution (Hindi)

Visualizing the Setup

Box 1: ${1, 2, 3}$ (Total 3 cards)
Box 2: ${1, 2, 3, 4, 5}$ (Total 5 cards)
Box 3: ${1, 2, 3, 4, 5, 6, 7}$ (Total 7 cards)
Total Outcomes: $3 \times 5 \times 7 = 105$

Odd and Even Probabilities

Box 1: $P (O_{1}) = \frac{2}{3}, P (E_{1}) = \frac{1}{3}$
Box 2: $P (O_{2}) = \frac{3}{5}, P (E_{2}) = \frac{2}{5}$
Box 3: $P (O_{3}) = \frac{4}{7}, P (E_{3}) = \frac{3}{7}$

Condition for Odd Sum

Sum $x_{1} + x_{2} + x_{3}$ is odd if:
1. $(O, O, O)$
2. $(O, E, E)$
3. $(E, O, E)$
4. $(E, E, O)$

Calculating Odd Sum Probability

$P (S u m i s o dd) = P (O_{1} O_{2} O_{3}) + P (O_{1} E_{2} E_{3}) + P (E_{1} O_{2} E_{3}) + P (E_{1} E_{2} O_{3})$
$= (\frac{2}{3} \cdot \frac{3}{5} \cdot \frac{4}{7}) + (\frac{2}{3} \cdot \frac{2}{5} \cdot \frac{3}{7}) + (\frac{1}{3} \cdot \frac{3}{5} \cdot \frac{3}{7}) + (\frac{1}{3} \cdot \frac{2}{5} \cdot \frac{4}{7})$
$= \frac{24 + 12 + 9 + 8}{105} = \frac{53}{105}$

Condition for Arithmetic Progression

$x_{1}, x_{2}, x_{3}$ are in AP if $2 x_{2} = x_{1} + x_{3}$
This implies $x_{1} + x_{3}$ must be even.
Also, $x_{2} = \frac{x _{1} + x _{3}}{2}$ must be an integer in ${1, 2, 3, 4, 5}$ .

Counting Favorable AP Cases

Possible $(x_{1}, x_{3})$ pairs where $x_{1} + x_{3}$ is even:
Sum = 2: $(1, 1) ⟹ x_{2} = 1$ (1 case)
Sum = 4: $(1, 3), (2, 2), (3, 1) ⟹ x_{2} = 2$ (3 cases)
Sum = 6: $(1, 5), (2, 4), (3, 3) ⟹ x_{2} = 3$ (3 cases)
Sum = 8: $(1, 7), (2, 6), (3, 5) ⟹ x_{2} = 4$ (3 cases)
Sum = 10: $(3, 7) ⟹ x_{2} = 5$ (1 case)

Final AP Probability

Total Favorable Cases = $1 + 3 + 3 + 3 + 1 = 11$
Total Outcomes = $105$
$P (A P) = \frac{11}{105}$

Summary and Key Takeaways

Key Takeaway 1: For independent events, multiply individual probabilities.
Key Takeaway 2: Sum of 3 numbers is odd in 4 specific parity combinations.
Key Takeaway 3: $a, b, c$ in AP $⟹ a + c = 2 b$ (Sum of ends must be even).
Next Challenge: What if we draw cards without replacement from a single large box?

00:00 / 00:00

Conceptually Similar Problems

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Comprehension Passage

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Let

n_{1}

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n_{2}

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n_{4}

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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)

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(C)

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(D)

n_{1} = 3

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Comprehension Passage

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Visualized Solution (Hindi)

Visualizing the Setup

Odd and Even Probabilities

Condition for Odd Sum

Calculating Odd Sum Probability

Condition for Arithmetic Progression

Counting Favorable AP Cases

Final AP Probability

Summary and Key Takeaways

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