In spite of all great answers, given here, I say, why not give another proof, from another point of view. The problem is we have 10 random variables $X_i$ for $i=1,\dots,10$, defined over $[6]=\{1,\dots,6\}$, and we are interested in distribution of $Z$ defined as $$ Z=X_1\oplus X_2\oplus \dots \oplus X_{10} $$ where $\oplus$ is addition modulo $6$. We can go on by two different, yet similar proofs. Show First proof: If $X_1$ and $X_2$ are two random variables over $[6]$, and $X_1$ is uniformly distributed, sheer calculation can show that $X_1\oplus X_2$ is also uniformly distributed. Same logic yields that $Z$ is uniformly distributed over $[6]$. Remark: This proves a more general problem. It says that even if only one of the dices is fair dice, i.e. each side appearing with probability $\frac 16$, the distribution of $Z$ will be uniform and hence $\mathbb P(Z=0)=\frac 16$. Second proof: This proof draws on (simple) information theoretic tools and assumes its background. The random variable $Z$ is output of an additive noisy channel and it is known that the worst case is uniformly distributed noise. In other word if $X_i$ is uniform for only one $i$, $Z$ will be uniform. To see this, suppose that $X_1$ is uniformly distributed. Then consider the following mutual information $I(X_2,X_3,\dots,X_6;Z)$ which can be written as $H(Z)-H(Z|X_2,\dots,X_6)$. But we have: $$ H(Z|X_2,\dots,X_6)=H(X_1|X_2,\dots,X_6)=H(X_1) $$
Similarly here, only one fair dice is enough. Moreover the same proof can be used for an arbitrary set $[n]$. As long as one of the $X_i$'s is uniform, then their finite sum modulo $n$ will be uniformly distributed. What is the probability that the sum of two dice is 10?The probability that the sum of the number appearing is more than 10 is.
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. What is the probability that the sum of the dice is a 10?So that's three ways of making a 10 out of 36 ways of rolling the two dice (6 for the first times 6 for the second). That means the probability is 3/36 which reduces to 1/12 or 0.083.
What is the probability of rolling a 10 with two dice?Two (6-sided) dice roll probability table. What is the probability that the sum of the numbers on both the dice is divisible by4 or 6?Required probability = P(S=4 or 6)P(Total)=1436=718.
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