My Math Forum Expected Total value

 Probability and Statistics Basic Probability and Statistics Math Forum

 June 10th, 2015, 03:14 PM #1 Newbie   Joined: Jun 2015 From: Kiribati Posts: 5 Thanks: 0 Math Focus: Statistics Expected Total value Could anyone check my working out Two coins are selected t random from a wallet containing four 5 cent pieces, one 10 cent piece, three 20 cent pieces and two 50 cent pieces. What is the expected total value of the two coins? expected value = mean = ∑(x).P(x) But before using the formula you should created your Probability distribution table: cents 5c 10c 20c 50c Probability distribution table value of x 4 1 3 2 then X 1 2 3 4 P(x=x) 2/5 1/10 3/10 1/5 P(X=x) 0.1 0.2 0.3 0.4 By using the formula(∑(x).P(x)) all x values are multiplied by its probability and then add them together. = 3 I dont know if I do the right thing because and if I do some mistakes I would like to learn from it.
June 10th, 2015, 05:21 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

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Hello, sensational!

Quote:
 Two coins are selected a random from a wallet containing four 5-cent pieces, one 10-cent piece, three 20-cent pieces and two 50-cent pieces. What is the expected total value of the two coins?

$\text{There are: }\,{10\choose2}\,=\,45\,\text{ pssible pairs of coins.}$

$\begin{array}{cccccccccccc} \text{Pair} &&\text{Prob}&&&& \text{Value} && \text{Exp} \\ \hline
(5,5) && \frac{{4\choose2}}{45} &=& \frac{6}{45} && 10 && \frac{60}{45} \\
(5,10) && \frac{{4\choose1}{1\choose1}}{45} &=& \frac{4}{45} && 15 && \frac{60}{45} \\
(5,20) && \frac{{4\choose1}{3\choose1}}{45} &=& \frac{12}{45} && 25 && \frac{300}{45} \\
(5,50) && \frac{{4\choose1}{2\choose1}}{45} &=& \frac{8}{45} && 55 && \frac{440}{45} \\
(10,20) && \frac{{1\choose1}{3\choose1}}{45} &=& \frac{3}{45} && 30 && \frac{90}{45} \\
(10,50) && \frac{{1\choose1}{2\choose1}}{45} &=& \frac{2}{45} && 60 && \frac{120}{45} \\
(20,20) && \frac{{3\choose2}}{45} &=& \frac{3}{45} && 40 && \frac{120}{45} \\
(20,50) && \frac{{3\choose1}{2\choose1}}{45} &=& \frac{6}{45} && 70 && \frac{420}{45} \\
(50,50) && \frac{{2\choose2}}{45} &=& \frac{1}{45} && 100 && \frac{100}{45} \\ \hline
&&&&&& \text{Total:} && \frac{1710}{45}\end{array}$

$E \;=\;\frac{1710}{45} \;=\;38\text{ cents.}$

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### two coins are selected at random from a wallet containing 1 five cent piece, 1 ten cent piece

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