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frez [133]
1 year ago
14

In a state's Pick 3 lottery game, you pay $1.35 to select a sequence of three digits (from 0 to 9), such as 366. If you select t

he same sequence of three digits that are drawn, you win and collect $477.75. Find the expected value.
Mathematics
1 answer:
tia_tia [17]1 year ago
5 0

Let us first review what we are given in the problem:

the game costs $1.35

3 digits are to be selected from 0 to 9

repetition is allowed (based on the given example)

a winning sequence earns $477.75

a. How many different selections are possible?

There are three slots and each slot has 10 digits to choose from. So there are 10 x 10 x 10 = 1,000 possible selections (from 000 up to 999)

b. What is the probability of winning?

The probablity of winning is equal to the number of ways you can win, divided by the total number of possible selections. Since there is ony one sequence that will win, then the answer is 1/1,000 or 0.001.

c. If you win, what is your net profit?

If you win, then your profit is equal to $477.75 minus the cost of joining the game, which is $1.35. So your profit is $476.40.

Note that if you do not win, then you lose the $1.35 you paid for joining the game.

d. Find the expected value.

The expected value is equal to the summation of all the possible outcomes (profits or losses) multiplied by their corresponding probabilities. So we calculate it as follows:

\begin{gathered} E(x)=\sum_^xP(x) \\  \\ E(x)=476.40(0.001)+(-1.35)(1-0.001) \\ E(x)=476.40(0.001)+(-1.35)(0.999) \\ E(x)=-0.87225 \end{gathered}

The expected value is -0.87 or a loss of $0.87.

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A concert hall has 8,000 seats and two categories of ticket prices, $29 and $34. Assume that all seats in each category can be s
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Using a system of equations, it is found that:

  • For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.
  • For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.
  • For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

--------------------

The variables of the system are:

  • x, which is the number of $29 tickets sold.
  • y, which is the number of $34 tickets sold.

Total of 8,000 seats, all can be sold, thus:

x + y = 8000 \rightarrow x = 8000 - y

--------------------

For a profit of $255,000, we have that:

29x + 34y = 255000

29(8000 - y) + 34y = 255000

5y = 23000

y = \frac{23000}{5}

y = 4600

x = 8000 - 4600 = 3400

For a profit of $255,000, 3400 tickets of $29 and 4600 tickets of $34 must be sold.

--------------------

For a profit of $271,000, we have that:

29x + 34y = 271000

29(8000 - y) + 34y = 271000

5y = 39000

y = \frac{39000}{5}

y = 7800

x = 8000 - 7800= 200

For a profit of $271,000, 200 tickets of $29 and 7800 tickets of $34 must be sold.

--------------------

For a profit of $235,000, we have that:

29x + 34y = 235000

29(8000 - y) + 34y = 235000

5y = 3000

y = \frac{3000}{5}

y = 600

x = 8000 - 600 = 7400

For a profit of $235,000, 7400 tickets of $29 and 600 tickets of $34 must be sold.

A similar problem is given at brainly.com/question/22826010

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