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If two different numbers are taken from the set {0,1,2,3, ......, 10} then what is the probability that their sum as well as absolute difference are both multiples of 4

Here is my work out

The sample space here is equal to 55.

Now to me the possible combinations are {0,4},{0,8},{2,6},{2,10},{4,8},{6,10}

so to me the answer is 6/55

Pole_Star
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3 Answers3

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Let's think what it means for two numbers $a, b$ to have sum and difference multiples of 4.

$a+b\equiv0 \pmod4$

$a-b\equiv0 \pmod4$

Adding them up, $2a\equiv0 \pmod4$, so it follows that $a\equiv0,2 \pmod4$

$a-b\equiv0 \pmod4$ states that $a\equiv b \pmod4$.

So, $a, b$ are $0, 2 \pmod4$ and are congruent to each other. This splits nicely into 2 cases, $0 \pmod4$ and $2 \pmod4$.

Under the case $0 \pmod4$, $a, b$ can be $0, 4, 8$. There are 3 ways to select 2 numbers from this set, and it is easy to verify that all 3 ways work.

Under the case $2 \pmod4$, $a, b$ can be $2, 6, 10$. There are 3 ways to select 2 numbers from this set, and it is easy to verify that all 3 ways work.

With $6$ satifying possibilities in total, and ${11\choose2}=55$ ways to choose 2 numbers, the probability of this event is $\frac{6}{55}$.

Element118
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  • yeah this is a good one!(+1) – Pole_Star Apr 05 '17 at 16:43
  • @Element118: I considered the order of rhe numbers too- as in, considering 0,2 and 2,0 as different cases. You still get the same answer because the 2s cancel out. Shouldn't that be how it's done? – harry Apr 15 '21 at 16:39
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    @HarryHolmes Both are equivalent, I think you can simply choose which method you are more comfortable with. – Element118 Apr 18 '21 at 06:19
  • @Element118: so if I'm in a situation where considering order just cancels out- for example say the numerator and denominator in the classical method both can have the data on the order factored out and cancelled out- then can I understand the situation to be one where order doesn't matter? And similarly that it does when it doesn't cancel out? Is that how you determine if you've got to consider order in probability? – harry Apr 18 '21 at 06:37
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    @HarryHolmes yes, if you see that order matters, then you have to consider the sample space as all possibilities, taking into account the order. – Element118 Apr 19 '21 at 11:21
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6/55 is correct sample space I {(0,4),(4,0),(0,8),(8,0),(2,6),(6,2),(2,10),(10,2),(4,8),(8,4),(6,10),(10,6)} terms are 12 And total number of ways we can arrange them is 11*10 Which is 110 now 12/110 is 6/55

Mohit
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Assuming random selection-- To have both sum and absolute difference of different integers be a multiple of 4 there are 5 possible combination of integers.0 and 4 , 2 and 6, 4 and 8, 6 and 10, and 0 and 8, = 5 choices Given that choice of 2 numbers from 0 to 10 , that is 11 choices, and that they must be different, the total choices is 11x10 = 110. The probability of the 5 combinations being chosen, remembering that we halve because there is no requirement for order, is 5 (times 2) divided by 110 = 1/11

John A S
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