Math / Data & Statistics

QuestionA certain game involves tossing 3 fair coins, and it pays $9 \notin$ for 3 heads, $4 \notin$ for 2 heads, fand $2 \notin$ for 1 head. Is $4 \notin$ a fair price to pay to play this game? That is, does the $4 \phi$ cost to play make the game fair?
The $4 c$ cost to play $\square$ is not a fair price to pay because the expected winnings are $\square$ 4. (Type an integer or a fraction. Simplify your answer.)

Studdy Solution

STEP 1

What is this asking? Is paying $4$ cents a fair price for a game where you toss three coins and win $\$9$ cents for three heads, $4$ cents for two heads, and $2$ cents for one head? Watch out! Don't forget to consider all the possible outcomes when calculating the expected winnings!

STEP 2

1. Calculate probability of each outcome
2. Calculate expected winnings for each outcome
3. Calculate total expected winnings

STEP 3

Let's start by figuring out the probability of getting three heads.
Since each coin flip is independent, we multiply the probabilities of getting a head on each individual flip.
The probability of a head on a single fair coin toss is $\frac{1}{2}$ .
So, the probability of three heads is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .

STEP 4

Now, for two heads.
There are three ways to get two heads and one tail: HHT, HTH, or THH.
Each of these has a probability of $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .
Since there are three such combinations, the total probability of getting two heads is $3 \cdot \frac{1}{8} = \frac{3}{8}$ .

STEP 5

Similarly, for one head, there are three ways: HTT, THT, or TTH.
Each has a probability of $\frac{1}{8}$ .
So, the total probability of getting one head is $3 \cdot \frac{1}{8} = \frac{3}{8}$ .

STEP 6

Finally, the probability of getting zero heads (all tails) is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .

STEP 7

The expected winnings for each outcome are calculated by multiplying the probability of that outcome by the amount you win for that outcome.

STEP 8

For three heads, the expected winnings are $\frac{1}{8} \cdot 9 = \frac{9}{8}$ cents.

STEP 9

For two heads, the expected winnings are $\frac{3}{8} \cdot 4 = \frac{12}{8}$ cents.

STEP 10

For one head, the expected winnings are $\frac{3}{8} \cdot 2 = \frac{6}{8}$ cents.

STEP 11

For zero heads, you win $\$0$ , so the expected winnings are $\frac{1}{8} \cdot 0 = 0$ cents.

STEP 12

To find the total expected winnings, we add up the expected winnings for each outcome: $\frac{9}{8} + \frac{12}{8} + \frac{6}{8} + 0 = \frac{9+12+6}{8} = \frac{27}{8}$ cents.

STEP 13

The expected winnings are $\frac{27}{8}$ cents, which is $3.375$ cents.
Since this is less than the $4$ cent cost to play, the $4$ cent cost is not a fair price.

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Studdy Solution

STEP 1

What is this asking? Is paying $4$ cents a fair price for a game where you toss three coins and win $\$9$ cents for three heads, $4$ cents for two heads, and $2$ cents for one head? Watch out! Don't forget to consider all the possible outcomes when calculating the expected winnings!

STEP 2

1. Calculate probability of each outcome
2. Calculate expected winnings for each outcome
3. Calculate total expected winnings

STEP 3

STEP 4

STEP 5

Similarly, for one head, there are three ways: HTT, THT, or TTH.
Each has a probability of $\frac{1}{8}$ .
So, the total probability of getting one head is $3 \cdot \frac{1}{8} = \frac{3}{8}$ .

STEP 6

Finally, the probability of getting zero heads (all tails) is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .

STEP 7

The expected winnings for each outcome are calculated by multiplying the probability of that outcome by the amount you win for that outcome.

STEP 8

For three heads, the expected winnings are $\frac{1}{8} \cdot 9 = \frac{9}{8}$ cents.

STEP 9

For two heads, the expected winnings are $\frac{3}{8} \cdot 4 = \frac{12}{8}$ cents.

STEP 10

For one head, the expected winnings are $\frac{3}{8} \cdot 2 = \frac{6}{8}$ cents.

STEP 11

For zero heads, you win $\$0$ , so the expected winnings are $\frac{1}{8} \cdot 0 = 0$ cents.

STEP 12

To find the total expected winnings, we add up the expected winnings for each outcome: $\frac{9}{8} + \frac{12}{8} + \frac{6}{8} + 0 = \frac{9+12+6}{8} = \frac{27}{8}$ cents.

STEP 13

The expected winnings are $\frac{27}{8}$ cents, which is $3.375$ cents.
Since this is less than the $4$ cent cost to play, the $4$ cent cost is not a fair price.

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Studdy Solution

STEP 1

What is this asking? Is paying 444 cents a fair price for a game where you toss three coins and win $9\$9$9 cents for three heads, 444 cents for two heads, and 222 cents for one head? Watch out! Don't forget to consider all the possible outcomes when calculating the expected winnings!

STEP 2

1. Calculate probability of each outcome2. Calculate expected winnings for each outcome3. Calculate total expected winnings

STEP 3

STEP 4

STEP 5

Similarly, for one head, there are three ways: HTT, THT, or TTH.Each has a probability of 18\frac{1}{8}81​.So, the total probability of getting one head is 3⋅18=383 \cdot \frac{1}{8} = \frac{3}{8}3⋅81​=83​.

STEP 6

Finally, the probability of getting zero heads (all tails) is 12⋅12⋅12=18\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}21​⋅21​⋅21​=81​.

STEP 7

The **expected winnings** for each outcome are calculated by multiplying the probability of that outcome by the amount you win for that outcome.

STEP 8

For three heads, the expected winnings are 18⋅9=98\frac{1}{8} \cdot 9 = \frac{9}{8}81​⋅9=89​ cents.

STEP 9

For two heads, the expected winnings are 38⋅4=128\frac{3}{8} \cdot 4 = \frac{12}{8}83​⋅4=812​ cents.

STEP 10

For one head, the expected winnings are 38⋅2=68\frac{3}{8} \cdot 2 = \frac{6}{8}83​⋅2=86​ cents.

STEP 11

For zero heads, you win $0\$0$0, so the expected winnings are 18⋅0=0\frac{1}{8} \cdot 0 = 081​⋅0=0 cents.

STEP 12

To find the **total expected winnings**, we add up the expected winnings for each outcome: 98+128+68+0=9+12+68=278\frac{9}{8} + \frac{12}{8} + \frac{6}{8} + 0 = \frac{9+12+6}{8} = \frac{27}{8}89​+812​+86​+0=89+12+6​=827​ cents.

STEP 13

The expected winnings are 278\frac{27}{8}827​ cents, which is 3.3753.3753.375 cents.Since this is less than the 444 cent cost to play, the 444 cent cost is *not* a fair price.

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What is this asking? Is paying $4$ cents a fair price for a game where you toss three coins and win $\$9$ cents for three heads, $4$ cents for two heads, and $2$ cents for one head? Watch out! Don't forget to consider all the possible outcomes when calculating the expected winnings!

1. Calculate probability of each outcome
2. Calculate expected winnings for each outcome
3. Calculate total expected winnings

Similarly, for one head, there are three ways: HTT, THT, or TTH.
Each has a probability of $\frac{1}{8}$ .
So, the total probability of getting one head is $3 \cdot \frac{1}{8} = \frac{3}{8}$ .

Finally, the probability of getting zero heads (all tails) is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$ .

The expected winnings for each outcome are calculated by multiplying the probability of that outcome by the amount you win for that outcome.

For three heads, the expected winnings are $\frac{1}{8} \cdot 9 = \frac{9}{8}$ cents.

For two heads, the expected winnings are $\frac{3}{8} \cdot 4 = \frac{12}{8}$ cents.

For one head, the expected winnings are $\frac{3}{8} \cdot 2 = \frac{6}{8}$ cents.

For zero heads, you win $\$0$ , so the expected winnings are $\frac{1}{8} \cdot 0 = 0$ cents.

To find the total expected winnings, we add up the expected winnings for each outcome: $\frac{9}{8} + \frac{12}{8} + \frac{6}{8} + 0 = \frac{9+12+6}{8} = \frac{27}{8}$ cents.

The expected winnings are $\frac{27}{8}$ cents, which is $3.375$ cents.
Since this is less than the $4$ cent cost to play, the $4$ cent cost is not a fair price.