JEE Challenger
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MathMedAdv 2014 • Paper 2

Probability that Numbers Drawn From Three Boxes Form an Arithmetic Progression

Comprehension Passage

Suppose Box 1 holds 3 cards numbered 1, 2, 3; Box 2 holds 5 cards labeled 1, 2, 3, 4, 5; and Box 3 holds 7 cards labeled 1, 2, 3, 4, 5, 6, 7. One card is drawn randomly from each box. Let xix_i denote the number on the card selected from the ithi^{\text{th}} box, where i=1,2,3i = 1, 2, 3.

Determine the probability that x1,x2,x3x_1, x_2, x_3 form an arithmetic progression in that sequence.

Options

A

9105\frac{9}{105}

B

10105\frac{10}{105}

C

11105\frac{11}{105}

Correct
D

7105\frac{7}{105}

Topics & Concepts

Probability (Advanced)ProbabilityArithmetic Progression

Step-by-Step Solution

To find the probability that x1,x2,x3x_1, x_2, x_3 form an arithmetic progression (A.P.) in that given sequence, we need to find the number of favorable outcomes where x1,x2,x3x_1, x_2, x_3 satisfy the definition of an A.P.

1. Sample Space Calculation

The cards are drawn independently from three boxes:

  • Box 1: x1∈{1,2,3}x_1 \in \{1, 2, 3\} (3 possibilities)
  • Box 2: x2∈{1,2,3,4,5}x_2 \in \{1, 2, 3, 4, 5\} (5 possibilities)
  • Box 3: x3∈{1,2,3,4,5,6,7}x_3 \in \{1, 2, 3, 4, 5, 6, 7\} (7 possibilities)

The total number of possible outcomes (sample space n(S)n(S)) is: n(S)=3×5×7=105n(S) = 3 \times 5 \times 7 = 105

2. Condition for Arithmetic Progression

For x1,x2,x3x_1, x_2, x_3 to be in an arithmetic progression, the middle term x2x_2 must be the average of x1x_1 and x3x_3: 2x2=x1+x3  ⟹  x3=2x2−x12x_2 = x_1 + x_3 \implies x_3 = 2x_2 - x_1

Since x3x_3 must belong to Box 3, it must satisfy 1≤x3≤71 \le x_3 \le 7, which gives: 1≤2x2−x1≤71 \le 2x_2 - x_1 \le 7

We analyze the possible values of x2x_2 and x3x_3 for each possible value of x1x_1:

Case 1: x1=1x_1 = 1

x3=2x2−1x_3 = 2x_2 - 1 We require 1≤2x2−1≤7  ⟹  2≤2x2≤8  ⟹  1≤x2≤41 \le 2x_2 - 1 \le 7 \implies 2 \le 2x_2 \le 8 \implies 1 \le x_2 \le 4.

  • If x2=1  ⟹  x3=1x_2 = 1 \implies x_3 = 1
  • If x2=2  ⟹  x3=3x_2 = 2 \implies x_3 = 3
  • If x2=3  ⟹  x3=5x_2 = 3 \implies x_3 = 5
  • If x2=4  ⟹  x3=7x_2 = 4 \implies x_3 = 7

This gives 4 valid triples: (1,1,1)(1, 1, 1), (1,2,3)(1, 2, 3), (1,3,5)(1, 3, 5), and (1,4,7)(1, 4, 7).

Case 2: x1=2x_1 = 2

x3=2x2−2x_3 = 2x_2 - 2 We require 1≤2x2−2≤7  ⟹  3≤2x2≤9  ⟹  1.5≤x2≤4.51 \le 2x_2 - 2 \le 7 \implies 3 \le 2x_2 \le 9 \implies 1.5 \le x_2 \le 4.5. Since x2x_2 is an integer, x2∈{2,3,4}x_2 \in \{2, 3, 4\}.

  • If x2=2  ⟹  x3=2x_2 = 2 \implies x_3 = 2
  • If x2=3  ⟹  x3=4x_2 = 3 \implies x_3 = 4
  • If x2=4  ⟹  x3=6x_2 = 4 \implies x_3 = 6

This gives 3 valid triples: (2,2,2)(2, 2, 2), (2,3,4)(2, 3, 4), and (2,4,6)(2, 4, 6).

Case 3: x1=3x_1 = 3

x3=2x2−3x_3 = 2x_2 - 3 We require 1≤2x2−3≤7  ⟹  4≤2x2≤10  ⟹  2≤x2≤51 \le 2x_2 - 3 \le 7 \implies 4 \le 2x_2 \le 10 \implies 2 \le x_2 \le 5.

  • If x2=2  ⟹  x3=1x_2 = 2 \implies x_3 = 1
  • If x2=3  ⟹  x3=3x_2 = 3 \implies x_3 = 3
  • If x2=4  ⟹  x3=5x_2 = 4 \implies x_3 = 5
  • If x2=5  ⟹  x3=7x_2 = 5 \implies x_3 = 7

This gives 4 valid triples: (3,2,1)(3, 2, 1), (3,3,3)(3, 3, 3), (3,4,5)(3, 4, 5), and (3,5,7)(3, 5, 7).

3. Total Favorable Outcomes and Probability

The total number of favorable outcomes n(E)n(E) is: n(E)=4+3+4=11n(E) = 4 + 3 + 4 = 11

Thus, the probability that x1,x2,x3x_1, x_2, x_3 form an arithmetic progression is: P(E)=n(E)n(S)=11105P(E) = \frac{n(E)}{n(S)} = \frac{11}{105}

Correct Option: C

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