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3 dice are rolled simultaneously. Event A: sum is multiple of 3 and event B: sum is multiple of 5.

Are this events independent?

There are 216 points in the sample space for this problem. Is there any way to do it without writing all points.

If they are independent or dependent..is there any intuitive way to reason about that?

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If $A$ and $B$ are independent, $P(A\text{ and }B)=P(A)\cdot P(B)$.

We need to convolve the distribution for one die, $\mathcal{P}=\left(\frac16,\frac16,\frac16,\frac16,\frac16,\frac16\right)$, with itself a couple of times. $$ \mathcal{P}\ast\mathcal{P}\ast\mathcal{P}=\left(\vphantom{\tfrac1{216}}\right.\overbrace{\tfrac1{216}}^3,\overbrace{\tfrac3{216}}^4,\overbrace{\tfrac6{216}}^5,\overbrace{\tfrac{10}{216}}^6,\overbrace{\tfrac{15}{216}}^7,\overbrace{\tfrac{21}{216}}^8,\overbrace{\tfrac{25}{216}}^9,\overbrace{\tfrac{27}{216}}^{10},\overbrace{\tfrac{27}{216}}^{11},\overbrace{\tfrac{25}{216}}^{12},\overbrace{\tfrac{21}{216}}^{13},\overbrace{\tfrac{15}{216}}^{14},\overbrace{\tfrac{10}{216}}^{15},\overbrace{\tfrac6{216}}^{16},\overbrace{\tfrac3{216}}^{17},\overbrace{\tfrac1{216}}^{18}\left.\vphantom{\tfrac1{216}}\right) $$ $P(\text{multiple of }3) =P(3)+P(6)+P(9)+P(12)+P(15)+P(18)=\frac{1+10+25+25+10+1}{216}=\frac13$

$P(\text{multiple of }5) =P(5)+P(10)+P(15)=\frac{6+27+10}{216}=\frac{43}{216}$

$P(\text{multiple of }15) =P(15)=\frac{10}{216}=\frac{5}{108}$

$P(\text{multiple of }3)\cdot P(\text{multiple of }5)=\frac{43}{648}$

Since $\frac{5}{108}\ne\frac{43}{648}$, the two events are not independent.

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A sum will be a multiple of both 3 and 5 if it is a multiple of 15.

Let $A$ be the event of a sum being a multiple of 3.   How many of the $6^3$ outcomes match this event?

Let $B$ be the event of a sum being a multiple of 5.   How many of the $6^3$ outcomes match this event?

Then $A\cap B$ be the event of a sum being a multiple of 15.   How many of the $6^3$ outcomes match this event?

You now have $\mathsf P(A), \mathsf P(B), \mathsf P(A\cap B)$.   If the events $A, B$ are independent, what should we observe?   Do we?

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Comment (already answered): A simulation of a million 3-roll experiments in R statistical software (correct to about three places) confirms the exact probabilities in @robjohn's Answer (+1). Thus according to @GrahamKemp's answer (+1), the events are not independent.

m = 10^6;  n = 3
f = sample(1:6, m*n, rep=T)
DTA = matrix(f, nrow=m) # each row has faces on 3 rolls
s = rowSums(DTA)  # vector of m sums of 3
pA = mean((s %% 3)==0)  # congruent to 0 (mod 3): divisible by 3
pB = mean((s %% 5)==0)
pAB = mean((s %% 15)==0)
pA;  pB; pAB; pA*pB
## 0.333622  # aprx P(div by 3)
## 0.199008  # aprx P(div by 5)
## 0.046195  # aprx P(div by 15)
## 0.06639345 # aprx product
1/3; 43/216; 5/108; 43/648  # exact fractions
## 0.3333333
## 0.1990741
## 0.0462963
## 0.06635802

The histogram below shows the simulated probabilities of sums (from 3 through 18). Magenta for divisible by 3; cyan for divisible by 5; blue for divisible by 15.

enter image description here

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