Great Expectations
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What is the expected sum of the numbers that come up when you roll a fair 6-sided die and a fair 12-sided die?
[Assume the faces have values 1-6 and 1-12, respectively.]The expectation of a sum is the sum of the individual expectations, so\(\frac{1+2+...+6}{6} + \frac{1+2+...+12}{12} = \frac{7}{2} + \frac{13}{2}=10\). -
Suppose you have two computers: Computer 1 generates a random number in the set \(\{1,2,\ldots,99\}\) with all numbers equally likely. Similarly, Computer 2 generates a random number in \(\{1,2,\ldots,999\}\) with all numbers equally likely.
You roll a fair die, and if a 5 comes up, you generate a random number using Computer 1, otherwise you generate a random number using Computer 2. What is the expected value of the number you generate?
Let \(R\) denote the value on the die, let \(G\) denote the generated number, let \(C_1\) denote the number Computer 1 generates and \(C_2\) the one Computer 2 generates.
By the law of Total Expectation, \[E[G] = E[G\;|\;R=5]\Pr[R=5] + E[G\;|\;R\neq 5]\Pr[R\neq 5]\] \(E[G\;|\;R=5] = E[C_1]=50\), since each number is equally likely. Similarly, \(E[G\;|\;R\neq 5]= E[C_2]=500\). Hence, \(E[G] = \frac{1}{6}\cdot 50+\frac{5}{6}\cdot 500 = \frac{2550}{6} = \fbox{425}.\) -
Assuming that Computers 1 & 2 act independently, what is the expected value of the product of the numbers they generate?
The Product Rule can be used because the two random variables are independent, so \[E[C_1\cdot C_2]=E[C_1]\cdot E[C_2]=50\cdot 500 = 25000.\]