Respuesta :
A. How many different selections are possible?
B. What is the probability of winning?
C. If you win, what is your net profit?
D. Find the expected value
Answer :
A. sequence of three digits (from 0 to 9). so 10 numbers.
We know 3 lottery games. So possible selections = 10^3 = 1000
B. Probability of winning = one outcome / total outcomes
P( winning) = [tex] \frac{1}{1000} [/tex] = 0.001
c. Net profit = winning amount - paid amount
= 384.13 - 1.33 = $382.8
D. expected value
E(x) = net profit * winning probability - paid amount * lost probability
E(x) = 382.8 * 0.001 - 1.33 * 0.999 = -0.95
you expect to lose 95 cents per ticket.
The expected value, sample size and net winning of the experiment in question is given below :
- Number of possible selections = 1000
- Probability of winning = 0.001
- Net winning = $382.80
- Expected value = - 0.945
The number of values that can be selected = (0 - 9) = 10 values
The number of possible selections
- [tex]10^{n} [/tex] ; n = Number of values to be selected
Number of ’possible selections = [tex]10^{3} = 1000[/tex]
The probability of winning :
- Number of winnings / Number of possible selections
Number of winnings = 1
Probability of winning = 1 / 1000 = 0.001
The net winning :
Winning amount - amount paid
$384.13 - $1.33 = $382.80
The Expected value :
(Net winning × winning probability) - (Amount paid × probability of not winning)
(382.80 × 0.001) - (1.33 × (1 - 0.001))
0.3282 - 1.32867 = - 0.945
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