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Probability question

2014 · Shift 2 · Q24
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  5. /2014 · Shift 2 · Q24

Probability question

2014 · Shift 2 · Q24

JEE AdvancedMathematicsProbabilityMCQ+3 / −1
Box 111 contains three cards bearing numbers 1,2,3;1,2,3;1,2,3; box 222 contains five cards bearing numbers 1,2,3,4,5;1,2,3,4,5;1,2,3,4,5; and box 333 contains seven cards bearing numbers 1,2,3,4,5,6,7.1,2,3,4,5,6,7.1,2,3,4,5,6,7. A card is drawn from each of the boxes. Let xi{x_i}xi​ be number on the card drawn from the ith{i^{th}}ith box, i=1,2,3.i=1,2,3.i=1,2,3. The probability that x1,{x_1},x1​,, x2,x3{x_2},{x_3}x2​,x3​ are in an arithmetic progression, is
  1. A
    9105{{9} \over {105}}1059​
  2. B
    10105{{10} \over {105}}10510​
  3. C
    11105{{11} \over {105}}10511​
  4. D
    7105{{7} \over {105}}1057​
View written solutionFree

Correct answer: C

  1. Total number of outcomes

From the three boxes, the numbers drawn are:

  • x1∈{1,2,3}x_1 \in \{1,2,3\}x1​∈{1,2,3}
  • x2∈{1,2,3,4,5}x_2 \in \{1,2,3,4,5\}x2​∈{1,2,3,4,5}
  • x3∈{1,2,3,4,5,6,7}x_3 \in \{1,2,3,4,5,6,7\}x3​∈{1,2,3,4,5,6,7}

Hence total possible outcomes are 3×5×7=105.3 \times 5 \times 7 = 105.3×5×7=105.

  1. Condition for arithmetic progression

For x1,x2,x3x_1,x_2,x_3x1​,x2​,x3​ to be in arithmetic progression, 2x2=x1+x3.2x_2 = x_1 + x_3.2x2​=x1​+x3​.

We count the number of triples (x1,x2,x3)(x_1,x_2,x_3)(x1​,x2​,x3​) satisfying this.

  1. Check each possible value of x1x_1x1​

Case 1: x1=1x_1=1x1​=1

Then 2x2=1+x3  ⟹  x3=2x2−1.2x_2 = 1 + x_3 \implies x_3 = 2x_2 - 1.2x2​=1+x3​⟹x3​=2x2​−1. Now x2∈{1,2,3,4,5}x_2 \in \{1,2,3,4,5\}x2​∈{1,2,3,4,5}, so corresponding x3x_3x3​ values are:

  • x2=1⇒x3=1x_2=1 \Rightarrow x_3=1x2​=1⇒x3​=1
  • x2=2⇒x3=3x_2=2 \Rightarrow x_3=3x2​=2⇒x3​=3
  • x2=3⇒x3=5x_2=3 \Rightarrow x_3=5x2​=3⇒x3​=5
  • x2=4⇒x3=7x_2=4 \Rightarrow x_3=7x2​=4⇒x3​=7
  • x2=5⇒x3=9x_2=5 \Rightarrow x_3=9x2​=5⇒x3​=9 (not allowed)

Valid triples: (1,1,1),(1,2,3),(1,3,5),(1,4,7).(1,1,1), (1,2,3), (1,3,5), (1,4,7).(1,1,1),(1,2,3),(1,3,5),(1,4,7). So this case gives 4 outcomes.

Case 2: x1=2x_1=2x1​=2

Then 2x2=2+x3  ⟹  x3=2x2−2.2x_2 = 2 + x_3 \implies x_3 = 2x_2 - 2.2x2​=2+x3​⟹x3​=2x2​−2. Possible values:

  • x2=1⇒x3=0x_2=1 \Rightarrow x_3=0x2​=1⇒x3​=0 (not allowed)
  • x2=2⇒x3=2x_2=2 \Rightarrow x_3=2x2​=2⇒x3​=2
  • x2=3⇒x3=4x_2=3 \Rightarrow x_3=4x2​=3⇒x3​=4
  • x2=4⇒x3=6x_2=4 \Rightarrow x_3=6x2​=4⇒x3​=6
  • x2=5⇒x3=8x_2=5 \Rightarrow x_3=8x2​=5⇒x3​=8 (not allowed)

Valid triples: (2,2,2),(2,3,4),(2,4,6).(2,2,2), (2,3,4), (2,4,6).(2,2,2),(2,3,4),(2,4,6). So this case gives 3 outcomes.

Case 3: x1=3x_1=3x1​=3

Then 2x2=3+x3  ⟹  x3=2x2−3.2x_2 = 3 + x_3 \implies x_3 = 2x_2 - 3.2x2​=3+x3​⟹x3​=2x2​−3. Possible values:

  • x2=1⇒x3=−1x_2=1 \Rightarrow x_3=-1x2​=1⇒x3​=−1 (not allowed)
  • x2=2⇒x3=1x_2=2 \Rightarrow x_3=1x2​=2⇒x3​=1
  • x2=3⇒x3=3x_2=3 \Rightarrow x_3=3x2​=3⇒x3​=3
  • x2=4⇒x3=5x_2=4 \Rightarrow x_3=5x2​=4⇒x3​=5
  • x2=5⇒x3=7x_2=5 \Rightarrow x_3=7x2​=5⇒x3​=7

Valid triples: (3,2,1),(3,3,3),(3,4,5),(3,5,7).(3,2,1), (3,3,3), (3,4,5), (3,5,7).(3,2,1),(3,3,3),(3,4,5),(3,5,7). So this case gives 4 outcomes.

  1. Total favorable outcomes

Therefore total favorable triples are 4+3+4=11.4+3+4=11.4+3+4=11.

  1. Required probability

P(A.P.)=11105.P(\text{A.P.}) = \frac{11}{105}.P(A.P.)=10511​.

  1. Compare with options

Option C\textbf{C}C is 11105,\frac{11}{105},10511​, which matches our result.

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