3 Boxes

Probability Level 3

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 the number on the card drawn from the $i^{th}$ box, $i=1,2,3$

What is the probability that $x_1,x_2,x_3$ are in A.P. (Arithmetic progression)?

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

\dfrac {10}{105}

\dfrac {12}{105}

\dfrac {11}{105}

\dfrac {9}{105}

2 solutions

Jordan Cahn
Dec 7, 2018

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

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

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

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.

3 Boxes

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