Answer:
The expected value of the safe bet equal $0
Step-by-step explanation:
If
is a finite numeric sample space and
for k=1, 2,..., n
is its probability distribution, then the expected value of the distribution is defined as
What is the expected value of the safe bet?
In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is
S = {100,-100}
Since the coin is supposed to be fair,
P(X=100)=0.5
P(X=-100)=0.5
and the expected value is
E(X) = 100*0.5 - 100*0.5 = 0
Without rounding to the nearest tenth, it would be 7.28. Rounding it would give you with an answer of 7.3 with rounding.
Both denominators are 4, so we can add the numerators to place over the common denominator. The numerators are -3 and -3, which add to -6. One way to think of negative numbers is to think of IOUs, which are a way of expressing debt in money. For instance, if you go into a store and buy a $10 item, but only have $3 in your pocket, then you would have to owe the owner $7. This can be represented with -7. If you repeat the process, then you'd have 7+7 = 14 in IOUs total. This would be represented with -14
In short, adding negative numbers is really the same as adding positives, but the final result is negative
So that's why -3+-3 turns into -6. We add the two threes like normal but then make the final result negative. All throughout this process, the denominator stays at 4.
So we end up with -6 over 4 which reduces to -3 over 2. How is this reduction happening? We are simply dividing each piece by the greatest common factor 2.
-6 divided by 2 = -3
4 divided by 2 = 2
The total amount that Shelley earned is $423.88.
Poop
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