4
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Arthur this is my workout...can u point how i can do it fast without devoting almost 2-3 minutes on this problem $\endgroup$
    – Pole_Star
    Apr 2, 2017 at 9:28
  • $\begingroup$ @Arthur workout shown $\endgroup$
    – Pole_Star
    Apr 2, 2017 at 9:29
  • $\begingroup$ Your solution is correct. $\endgroup$
    – Crostul
    Apr 2, 2017 at 9:34
  • $\begingroup$ is there any faster way to solve this problem? $\endgroup$
    – Pole_Star
    Apr 2, 2017 at 9:37

3 Answers 3

4
$\begingroup$

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 $\frac{6}{55}$.

$\endgroup$
5
  • $\begingroup$ yeah this is a good one!(+1) $\endgroup$
    – Pole_Star
    Apr 5, 2017 at 16:43
  • $\begingroup$ @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? $\endgroup$
    – harry
    Apr 15, 2021 at 16:39
  • 1
    $\begingroup$ @HarryHolmes Both are equivalent, I think you can simply choose which method you are more comfortable with. $\endgroup$
    – Element118
    Apr 18, 2021 at 6:19
  • $\begingroup$ @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? $\endgroup$
    – harry
    Apr 18, 2021 at 6:37
  • 1
    $\begingroup$ @HarryHolmes yes, if you see that order matters, then you have to consider the sample space as all possibilities, taking into account the order. $\endgroup$
    – Element118
    Apr 19, 2021 at 11:21
0
$\begingroup$

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

$\endgroup$
-2
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ the total number of choices is 11 choose 2 = 55 $\endgroup$
    – Harambe
    Apr 2, 2017 at 9:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .