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:
yea i can help
Step-by-step explanation:
So hmm the first term is -2
and if we divide one term by the term before it, we'd get the "common ratio" "r"
so hmm say -32/8 that gives us -4, so r = -4
thus
