Complete question :
Suppose someone gives you 8 to 2 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win $8 if you succeed and you lose $2 if you fail. What is the expected value of this game to you? What can you expect if you play 100 times.
Answer:
$0.5 ; win $50 with 100 rolls
Step-by-step explanation:
From a roll of two fair dice; probability of obtaining an even number :
Even numbers = (2, 4, 6) = 3
P = 3 /6 = 1 /2
For 2 fair dice ; probability of rolling two even numbers : independent event.
1/2 * 1/2 = 1/4
Hence, p(success) = 1/4 ; P(failure) = 1 - 1/4 = 3/4
Probability table
Winning = $8 or loss = - $2
X : ____ 8 ______ - 2
P(x) __ 1/4 ______ 3/4
Expected value : E(x) = ΣX*P(x)
E(x) = (8 * 1/4) + (-2 * 3/4)
E(x) = 2 - 1.5
E(x) = $0.5
Since expected value is positive, the expect to win
If played 100 times;
Expected value = 100 * $0.5 = $50
Answer:
x = 6
Step-by-step explanation:
<GHA + <AHI = <GHI
(8x + 2) + (13x + 9) = 137
8x + 13x + 2 + 9 = 137
21x= 137 - 11
x = 126/21
x = 6
Answer:
X = -10/6
X = -1,666666666666667
Step-by-step explanation:
First at all you join the X's
18x-12x = 14-24
Then you simplified:
6x = -10
X = -10/6
X = -1,666666666666667
Answer:
her balance aka ADB would be 372.08
12s - 20 < 3s - 25 |+20
12s < 3s - 5 |-3s
9s < -5 |:9
s < -5/9
Answer: Any number from the set (-∞, -5/9).
