Question

An um contains 1 red marble, 7 white marbles and 2 blue marbles. A player randomly
selects one marble from the urn and wins $5 if the marble is red, loses $3 if the marble
is white and loses $1 if the marble is blue. Find the expected winnings.

Answer 1

Answer:

A loss of 1.8$

Step-by-step explanation:

In the bag in this problem, we have:

r = 1 (number of red marbles)

w = 7 (number of white marbles)

b = 2 (number of blue marbles)

n = r + w + b = 1 + 7 + 2 = 10 (number of total marbles)

So, the probabilities of choosing a red, a white or a blue marble are:

$p(r) = \frac{r}{n}=\frac{1}{10}=0.1$ (probability of choosing a red marble)

$p(w)=\frac{w}{n}=\frac{7}{10}=0.7$ (probability of choosing a white marble)

$p(b) = \frac{b}{n}=\frac{2}{10}=0.2$ (probability of choosing a blue marble)

The winning associated to each probability are:

+$5 if the marble is red

-$3 if the marble  is white

-$1 if the marble is blue

So the expected winning can be calculated as follows:

$E(X)=p(r) \cdot (+\ 5)+p(w) \cdot (-\ 3) +p(b)\cdot (-\ 1)=\\=(0.1)(+5)+(0.7)(-3)+(0.2)(-1)=-\ 1.8$

So, he can expect a loss of 1.8$.