Question

A box contains nine $1 bills, eight $5 bills, two $10 bills, and four $20 bills. What is the expectation if one bill is selected? A.$6.48 B.$8.48 C.$5.48 D.$7.48

Answer 1

answer: A Guy With More Money Than Me.

Answer 2

lol that's good!!^^^^^

Answer 3

There's a total of 23 bills, with the following probabilities of selecting a particular denomination:
\[P($1)=\frac{9}{23}\\P($5)=\frac{8}{23}\\
P($10)=\frac{2}{23}\\P($20)=\frac{4}{23}\]
If \(X\) denotes the random variable for bill denomination, then \(X\) can take on any value from \(\{1,5,10,20\}\).

So, if \(x\) is one of the values of \(X\) and \(f(x)\) is the probability of a particular \(x\), the expected value is
\[E(X)=\sum_{\text{all }x}x~f(x)=1\cdot\frac{9}{23}+5\cdot\frac{8}{23}+10\cdot\frac{2}{23}+20\cdot\frac{4}{23}\approx6.48\]