Mathematics, 07.03.2020 01:25 joserodriguez212004

N a state's Pick 3 lottery game, you pay $1.14 to select a sequence of three digits (from 0 to 9), such as 688. If you select the same sequence of three digits that are drawn, you win and collect $336.22. Complete parts (a) through (e).

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Answer from: jazmine1260

Part a: This indicates that the total number of possible selection would be 1000.

Part b: The probability of winning is 0.001.

Part c: The net profit is $335.08.

Part d: The expected value is –$0.8026

Part e:The bet of 4 lottery game is better.

Step-by-step explanation:

As the complete question is not given, the complete question is found online and is as attached herewith.

Part a:

The different possible selection in three digits is obtained as shown below:

From the information, the three digits number is drawn. The lowest three digit number is 000 and highest three digit number is 999. The selection of three digit number lies between 000 and 999. The sample space of three digits number are given as,

$S = \left\{ {000,\,001,\,...,\,998,\,999} \right\}$

This indicates that the total number of possible selection would be 1000.

Part b:

The probability of winning is obtained as shown below:

Let the event A denotes the chance of winning.

From the information, the three digits number is drawn for winning. The total numbers of possible outcomes in three digits number are 1000.

The required probability is,

$\begin{array}{c}\\P\left( A \right) = \frac{{N\left( {{\rm{event}}} \right)}}{{N\left( S \right)}}\\\\ = \frac{1}{{1,000}}\\\\ = 0.001\\\end{array}$

The probability of winning is 0.001.

Part c:

The net profit is obtained as shown below:

From the information, the money for the same sequence of three digit number getting is $336.22 and the charge for selecting a three digit number is $1.14. The required net profit is,

$Profit = {Total\, winning\,cost} -{Total \,paid\, cost}\\= \$336.22-\$ 1.14\\=\$335.08$

The net profit is $335.08.

Part d:

The expected value is obtained as shown below:

From the given information, the cost for losing is Cost=$1.14 and the probability of losing is,

$\begin{array}{c}\\P\left( {{\rm{Losing}}} \right) = 1 - 0.001\\\\ = 0.999\\\end{array}$

and the cost for winning is Cost=$336.22 and the probability of winning is 0.001.

The required expected value is,

$\begin{array}{c}\\{\rm{Expected}} = \left( {{\rm{Cost}}\,{\rm{of winning}} \times {\rm{P}}\left( {{\rm{winning}}} \right)} \right)\, - \left( {{\rm{Cost}}\,{\rm{of}}\,{\rm{losing}} \times {\rm{P}}\left( {{\rm{Losing}}} \right)} \right)\\\\ = \left( {\$ 336.22 \times 0.001} \right) - \left( {\$ 1.14 \times 0.999} \right)\\\\ = \$0.33622-\$1.13886\\\\ = - \$ 0.8026\\\end{array}$

The expected value is –$0.8026

Part e:

From the information, the expected cost is –$1.01 in a 4 lottery game where as the expected value in 3 lottery game is –$0.80. The bet of 4 lottery game is better than the bet of 3 lottery game because the expected value of 4 lottery game is minimum as compared to 3 lottery game.

The bet of 4 lottery game is better.