3 Boxes

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 x_i be the number on the card drawn from the i t h i^{th} box, i = 1 , 2 , 3 i=1,2,3

What is the probability that x 1 , x 2 , x 3 x_1,x_2,x_3 are in A.P. (Arithmetic progression)?

Note :- If a , b , c a,b,c are in A.P., 2 b = a + c 2b=a+c

10 105 \dfrac {10}{105} 12 105 \dfrac {12}{105} 11 105 \dfrac {11}{105} 9 105 \dfrac {9}{105}

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2 solutions

Jordan Cahn
Dec 7, 2018

Using the note, we consider all possible values of x 2 x_2 and count the number of choices of x 1 x_1 and x 3 x_3 for which x 1 + x 3 = 2 x 2 x_1+x_3 = 2x_2

  • x 2 = 1 x_2=1 : There is one choice: 1 + 1 = 2 × 1 1+1=2\times 1
  • x 2 = 2 x_2=2 : There are three choices: 1 + 3 = 2 × 2 1+3=2\times 2 , 2 + 2 = 2 × 2 2+2 = 2\times 2 , 3 + 1 = 2 × 2 3 + 1 = 2\times 2
  • x 2 = 3 x_2=3 : There are three choices: 1 + 5 = 2 × 3 1+5=2\times 3 , 2 + 4 = 2 × 3 2+4 = 2\times 3 , 3 + 3 = 2 × 3 3 + 3 = 2\times 3
  • x 2 = 4 x_2=4 : There are three choices: 1 + 7 = 2 × 4 1+7=2\times 4 , 2 + 6 = 2 × 4 2+6 = 2\times 4 , 3 + 5 = 2 × 4 3 + 5 = 2\times 4
  • x 2 = 5 x_2=5 : There is one choice: 3 + 7 = 2 × 5 3+7=2\times 5

Thus there are 1 + 3 + 3 + 3 + 1 = 11 1+3+3+3+1 = 11 possible arithmetic progressions that can be formed. There are 3 × 5 × 7 = 105 3\times 5\times 7 = 105 total sequences, so the probability of drawing an arithmetic sequence is 11 105 \boxed{\dfrac{11}{105}} .

Jason Carrier
Dec 9, 2018

For a given pair of cards from the first two boxes, there is exactly one number that continues the arithmetic progression. There is a twist: not all of these numbers will land in the range of 1-7. Specifically, 1,5 and 2,5 give numbers too large, and 2,1 and 3,1 give numbers too small. So, 11/15 times it will be possible to draw the correct third card, at which point there will be a 1/7 chance of picking the right one. The overall chance is therefore 11/105.

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