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Solution by Steps

step 1

Identify the outcomes when rolling two dice. There are a total of 36 outcomes, as each die has 6 faces

step 2

Count the winning outcomes: You win $12 for rolling a double. The doubles are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), totaling 6 outcomes

step 3

Count the losing outcomes: You lose $6 if the sum is 7. The combinations that yield a sum of 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), totaling 6 outcomes

step 4

Calculate the remaining outcomes: The total outcomes are 36, and the outcomes that are either a double or sum to 7 are 12 (6 doubles + 6 sums of 7). Thus, the remaining outcomes are 36 - 12 = 24

step 5

Let

x

be the amount you win or lose for the remaining outcomes. To make the game fair, the expected value must equal zero:

\frac{6 \cdot 12 + 6 \cdot (-6) + 24 \cdot x}{36} = 0

step 6

Simplifying the equation:

72 - 36 + 24x = 0

leads to

36 + 24x = 0

step 7

Solving for

x

:

24x = -36

gives

x = -\frac{36}{24} = -\frac{3}{2} = -1.5

Answer

You should lose $1.50 if the sum is not 7 and the roll is not a double for the game to be fair.

Key Concept

Expected value in probability helps determine fair outcomes in games.

Explanation

The expected value calculation ensures that the game is balanced, meaning the average outcome over time will be zero, making it fair for both players.

Generated Graph

Solution by Steps

step 1

Let P(C) be the probability of having cancer, which is given as

P(C) = 0.02

(2% of 1000 adults)

step 2

Let P(T|C) be the probability of testing positive given that the person has cancer, which is

P(T|C) = 0.98

step 3

Let P(T|C') be the probability of testing positive given that the person does not have cancer, which is

P(T|C') = 0.01

step 4

The probability of not having cancer is

P(C') = 1 - P(C) = 0.98

step 5

We can use Bayes' theorem to find

P(C|T)

:

P(C|T) = \frac{P(T|C) \cdot P(C)}{P(T)}

where

P(T) = P(T|C) \cdot P(C) + P(T|C') \cdot P(C')

step 6

Calculate

P(T)

:

P(T) = (0.98 \cdot 0.02) + (0.01 \cdot 0.98) = 0.0196 + 0.0098 = 0.0294

step 7

Now substitute back into Bayes' theorem:

P(C|T) = \frac{0.98 \cdot 0.02}{0.0294} \approx 0.667

Answer

The probability that a randomly chosen person has cancer given that the test indicates cancer is approximately 0.667 or 66.7%.

Key Concept

Bayes' theorem is used to find conditional probabilities based on prior knowledge.

Explanation

The answer shows how to calculate the probability of having cancer given a positive test result, taking into account the accuracy of the test and the prevalence of the disease.

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