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QJEEAdv14Maa62c: JEE Advanced 2014 Mathematics - Class 12 JEEAdv14Maa62c Solution: Box contains three cards bearing numbers box contains five…

Box 11 contains three cards bearing numbers 1,2,3;1,2,3; box 22 contains five cards bearing numbers 1,2,3,4,5;1,2,3,4,5; and box 33 contains seven cards bearing numbers 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} be number on the card drawn from the ith{i^{th}} box, i=1,2,3.i=1,2,3.

The probability that x1,{x_1}, , x2,{x_2}, x3{x_3} are in an arithmetic progression, is

Multiple Choice Options

  • A 9105{{9} \over {105}}
  • B 10105{{10} \over {105}}
  • C 11105{{11} \over {105}}
  • D 7105{{7} \over {105}}

Step-by-Step Verified Solution

Given three Boxes :

JEE Advanced 2014 Paper 2 Offline Mathematics - Probability Question 36 English Explanation

Given, that a card is drow from each box and xix_i denote the number on the card drawn from ith i^{\text {th }} box, i=1,2,3i=1,2,3 .

Sample Space =3C1.5C1.7C1=3⋅5⋅7=105={ }^3 \mathrm{C}_1 .{ }^5 \mathrm{C}_1 .{ }^7 \mathrm{C}_1=3 \cdot 5 \cdot 7=105 .

Let B be the event when x1,x2,x3x_1, x_2, x_3 are in A.P.

⇒2x2=x1+x3\Rightarrow 2 x_2=x_1+x_3

Here, x1+x3x_1+x_3 is even number and for every even value of x1+x3x_1+x_3 there is only one possible value of x2x_2

x1+x2x_1+x_2 is even in two cases

Case I: When both x1x_1 and x2x_2 are even

No. of ways =1C1={ }^1 \mathrm{C}_1 . 3C1==3{ }^3 \mathrm{C}_{1=}=3

Case II: When both x1x_1 and x2x_2 are odd

No. of ways =2C1.4C1=2×4=8={ }^2 \mathrm{C}_1 .{ }^4 \mathrm{C}_1=2 \times 4=8

Number of favorable cases for B=3+8=11\mathrm{B}=3+8=11

Required probability =11105=\frac{11}{105}

Hint:

2x2=x1+x32 x_2=x_1+x_3 indicate, x1+x3x_1+x_3 is even and for every even value of x1+x3x_1+x_3 there is only one possible value of x2x_2 . Now, x1+x3x_1+x_3 is even in two cases:

(i) When both x1x_1 and x2x_2 are even.

(ii) When both x1x_1 and x2x_2 are odd.

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