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Expert-verifiedQuestion: What is the expected sum of the numbers that appear when three fair dice are rolled?
Answer:
The expected sum is\(10.5\).
Three fair dice are rolled.
The expected number of successes when\(n\) mutually independent Bernoulli trials are performed, where\(p\) is the probability of success on each trial, is \(np\).
Let S be the random variable denoting the sum of numbers when a fair die is rolled.
Let s1, s2, s3 denote the corresponding sums for the three dice.
We have \(S{\rm{ }} = {\rm{ }}s1 + s2 + s3\)
\(\begin{array}{l}E(s) = E(s1 + s2 + s3)\\ = E(s1) + E(s2) + E(s3)\end{array}\)
The expectation of the sum is the sum of the expectation values for three dice.
But since they all are fair; they all have equal expectation values.
Hence, \(E(s) = 3E(s1)\)
The outcomes for a single fair dice are\(1, 2, 3,4, 5 and 6\) with probability \(\frac{1}{6}\).
So, \(E(x) = \sum\limits_{i = 1}^6 {X.P(X = i)} \)
\(\begin{array}{l}E(s1) = \frac{{1 + 2 + 3 + 4 + 5 + 6}}{6}\\ = 3.5\\E(s) = 3 \times 3.5 = 10.5\end{array}\)
Thus, the expected sum is \(10.5\).
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