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
I believe your answer is -220w-22
2-3(2w+1)=7w-3(7+w)
+7w -7w
2-3(9w+1)=-3(7+w)
+w -w
2-3(10w+1)=-3(7)
2-3(10w+1)=-21
-1(10w+1=-21
-21 +21
-22(10w+1)
-220w-22
Answer:
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
eeeeeeeeeeeeeeeeeeeeeeStep-by-step explanation:
Answer: a. 1734 b. 477000
Step-by-step explanation:
We know that,
Distributive property : a(b+c) = ab+ac
a. 17 x 102
= 17 (100+2)
= 17 x 100 + 17 x 2 [Distributive property ]
= 1700+34
= 1734
b. 477×1035 -477×35
= 477 (1035-35) [Distributive property ]
= 477 (1000)
= 477000
3/5 cause you get the lowest common denominator
