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masha68 [24]
3 years ago
14

Three fair coins are tossed. If all land "heads," the player wins $10, and if exactly two land heads, the player wins $5. If it

costs $4 to play, what is the player's expected outcome after four games?
Mathematics
1 answer:
hodyreva [135]3 years ago
8 0

Answer:

<u>After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:</u>

<u>1/8 or 12.5% of landing three "heads"</u>

<u>3/8 or 37.5% of landing two "heads"</u>

<u>4/8 or 50% of landing no or only one "head".</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

If three coins land "heads" the player wins $ 10

If two coins land "heads" the player wins $ 5

Cost of playing = $ 4

2.  What is the player's expected outcome after four games?

Probability of two coins out of three lands "heads" = 3/8

Probability of three coins out of three lands "heads" = 1/8

Now, let's calculate the player's expected outcome, as follows:

Four games:

Cost = 4 * 4 = $ 16

Worst-case scenario: No wins

Best-case scenario: 4 out of 4 of $ 10 win

Worst-case scenario profit or loss = 0 - 16 = Loss of $ 16

Best-case scenario profit or loss = 40 - 16 = Profit of $ 24

After four games, a player can lose up to $ 16 to win up to $ 26. These are the probabilities for every game:

1/8 or 12.5% of landing three "heads"

3/8 or 37.5% of landing two "heads"

4/8 or 50% of landing no or only one "head".

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An expirement consists of rolling two fair number cubes. What is the probability that the sum of the two numbers will be 4? Expr
ExtremeBDS [4]

Answer:

\dfrac{1}{12}

Step-by-step explanation:

Given:

Two fair number cubes i.e. two dice consisting the numbers 1, 2, 3, 4, 5, 6 on their faces and have equal probability of each number.

The dice are rolled.

To find:

Probability of getting the sum of two numbers as 4.

Solution:

First of all, let us have a look at the total possibilities when two dice are rolled:

([1][1], [1][2], [1][3], [1][4], [1][5], [1][6],

[2][1], [2][2], [2][3], [2][4], [2][5], [2][6],

[3][1], [3][2], [3][3], [3][4], [3][5], [3][6],

[4][1], [4][2], [4][3], [4][4], [4][5], [4][6],

[5][1], [5][2], [5][3], [5][4], [5][5], [5][6],

[6][1], [6][2], [6][3], [6][4], [6][5], [6][6])

These are total 36 possible outcomes.

For getting a sum as 4:

Possible number of favorable cases are 3 (as highlighted in BOLD in above)

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

Required probability is:

\dfrac{3}{36} = \bold{\dfrac{1}{12}}

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Answer:

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Step-by-step explanation:

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So we can fill out the table, in every iteration wins increases by 7 and losses increases by 2.

When we fill this out, we find that when losses is 6, wins is 21, so when you have 2 losses you have 21 wins.

Hope this helped!

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Answer:

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Step-by-step explanation:

If you have one group of 832, and put it into 1 group you get 832

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