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Vaselesa [24]
2 years ago
5

A game consists of tossing three coins. If all three coins land on heads, then the player wins $75. If all three coins land on t

ails, then the player wins $45. Otherwise, the player wins nothing. On average, how much should a player expect to win each game
Mathematics
1 answer:
emmainna [20.7K]2 years ago
7 0

Answer:

On average, a player should expect to win $15.

Step-by-step explanation:

The expected value in an event with outcomes:

x₁, x₂, ..., xₙ

Each with probability:

p₁, ..., pₙ

is given by:

Ev = x₁*p₁ + ... +xₙ*pₙ

In this case we have 3 outcomes:

player wins $75 = x₁

player wins $45 = x₂

player does not win = x₃

Let's find the probabilities of these events.

player wins $75)

Here we must have the 3 coins landing on heads, so there is only one possible outcome to win $75

While the total number of outcomes for tossing 3 coins, is the product between the number of outcomes for each individual event (where the individual events are tossing each individual coin, each one with 2 outcomes)

Then the number total of outcomes is:

C = 2*2*2 = 8

Then the probability of winning $75 is the quotient between the number of outcomes to win (only one) and the total number of outcomes (8)

p₁ = 1/8

Win $45:

This happens if the 3 coins land on tails, so is exactly equal to the case above, and the probability is the same:

p₂ = 1/8

Not wining:

Remember that:

p₁ + p₂ + ... + pₙ = 1

Then for this case, we must have:

p₁ + p₂ + p₃ = 1

1/8 + 1/8 + p₃ = 1

p₃ = 1 - 1/8 - 1/8

p₃ = 6/8

Then the expected value will be:

Ev = $75*1/8 + $45*1/8 + $0*6/8 = $15

On average, a player should expect to win $15.

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alukav5142 [94]
I think that the answer is A.
3 0
2 years ago
I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.
masya89 [10]

Answer:

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 96.667

The variance is 22.222

b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 3.333

The variance is 33.333

Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

4 0
2 years ago
how man 1/3 inch cubes does it take to fill a box with width 2 2/3 inches,lenght 3 1/3 inches and height 2 1/3
galben [10]

idk anymore ;-; please WELP ME

3 0
3 years ago
Plz help I don't know how to salve the question
kvv77 [185]

Answer:

24

Step-by-step explanation:

4 0
2 years ago
A salesperson earns a commission based on the number and type of vehicle sold. A person selling 6 cars and 3 trucks earns $4,800
madam [21]
4800 divide by 3=1600 (2 cars 1 truck)

4600-1600-3000 (6 cars)

3000 divide by 3 = 1000(2cars)

1000 times 2=2000(4 cars )

4800-2000-2800 (2 cars 3 trucks)

$2,800 is your answer
8 0
2 years ago
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