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Chapter 5: Problem 99

Suppose a loaded die has the following probability model: $$\begin{array}{lcccccc}\hline \text { 0utcome: } & 1 & 2 & 3 & 4 & 5 & 6 \\\\\text { Probability: } & 0.3 & 0.1 & 0.1 & 0.1 & 0.1 & 0.3 \\\\\hline\end{array}$$ If this die is thrown and the top face shows an odd number, what is the probability that the die shows a l? \(\begin{array}{ll}\text { (a) } & 0.10\end{array}\) (b) 0.17 (c) 0.30 (d) 0.50 (e) 0.60

Short Answer

Expert verified
(e) 0.60

Step by step solution

01

Identify Odd Outcomes

The odd numbers on a die are 1, 3, and 5. According to the given probability model, these outcomes have probabilities: - P(1) = 0.3 - P(3) = 0.1 - P(5) = 0.1.
02

Calculate Total Probability of Odd Outcomes

We need to determine the total probability that the die shows an odd number. Add the probabilities of rolling a 1, 3, or 5: \[\text{Total Probability of Odd Outcomes} = P(1) + P(3) + P(5) = 0.3 + 0.1 + 0.1 = 0.5.\]
03

Probability of 1 Given Odd

Using conditional probability, calculate the probability that the outcome is 1 given it is odd. The formula for conditional probability \( P(A|B) = \frac{P(A \cap B)}{P(B)} \) applies here as:\[ P(1|\text{Odd}) = \frac{P(1)}{\text{Total Probability of Odd Outcomes}} = \frac{0.3}{0.5} = 0.6. \]
04

Find the Correct Answer Option

The calculated probability \( P(1|\text{Odd}) \) is 0.6. Therefore, the correct answer choice that corresponds to this probability is (e) 0.60.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Model
A probability model is like a blueprint that shows all possible outcomes of a random experiment along with their corresponding probabilities. It helps us understand how likely different events are.
For instance, rolling a die is a common experiment. In a fair die, each number has an equal chance of appearing. However, if the die is loaded鈥攎eaning it's biased鈥攕ome numbers are more likely to show up than others.
In our case, the loaded die shows that the probability of getting a 1 or a 6 is 0.3, while rolling a 2, 3, 4, or 5 has a much lower probability of 0.1 each. This model completely describes what can happen when you roll the die and precisely how often each result will occur.
Loaded Die
A loaded die is intentionally biased. This means it does not behave like a normal die.
In a standard die, each number, from 1 to 6, has an equal chance of being rolled. This chance is 1/6 or approximately 0.1667.
But in a loaded die, some outcomes are made more likely than others.
  • This might occur due to intentional changes in the die's weight or shape, skewing results in favor of certain numbers.
  • In our exercise, the probabilities are heavily loaded towards 1 and 6, both with a chance of 0.3, distorting the expected even spread of outcomes.
Understanding loaded dice can help in identifying unfair games or situations.
Outcomes
Outcomes in probability are simply the possible results that can occur from an experiment. For a die, the outcomes are the numbers 1 through 6.
However, the interesting part is when these outcomes are not equally likely because the die is loaded.
This changes how we look at the outcomes, as each number now carries a different weight or probability.
In our scenario, outcomes 1 and 6 are prime, each with a probability of 0.3, making them more common than the others with probabilities of 0.1. Observing these skewed results can be surprising. They influence decisions and provide insight into how the bias affects the game.
Step-by-Step Solution
Breaking down the exercise into step-by-step solutions makes it manageable and helps ensure understanding of each part.
Let's go over this particular problem:
  • **Step 1** was about identifying odd outcomes. These are specifically 1, 3, and 5, where we noted their respective probabilities.
  • **Step 2** involved calculating the total probability of getting an odd number. By summing the probabilities of these outcomes, we get 0.5.
  • **Step 3** applied the conditional probability formula to find the probability of rolling a 1, given the result was odd. This was calculated as 0.6.
  • **Step 4** simply required matching our calculated result to the correct answer choice, which was option (e) 0.60.
This structured approach not only helps in solving this particular problem but also offers a clear strategy for tackling similar problems in the future.

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Most popular questions from this chapter

Probability models? In each of the following situations, state whether or not the given assignment of probabilities to individual outcomes is legitimate, that is, satisfies the rules of probability. If not, give specific reasons for your answer. (a) Roll a 6-sided die and record the count of spots on the up-face: \(P(1)=0, P(2)=1 / 6, P(3)=1 / 3, P(4)=$$$1 / 3, P(5)=1 / 6, P(6)=0$$(b) Choose a college student at random and record gender and enrollment status: \)P(\( female full-time \))=0.56\(, \)P(\( male full-time \))=0.44, P(\( female part-time \))=0.24\( \)P(\( male part-time \))=0.17\( (c) Deal a card from a shuffled deck: \)P(\( clubs \))=12 / 52\(, \)P(\( diamond \)s)=12 / 52, P(\( heart \))=12 / 52, P(\( spades \))=\( \)16 / 52$

Genetics Suppose a married man and woman both carry a gene for cystic fibrosis but don't have the disease themselves. According to the laws of genetics, the probability that their first child will develop cystic fibrosis is 0.25 (a) Explain what this probability means. (b) If the couple has 4 children, is one of them guaranteed to get cystic fibrosis? Explain.

You read in a book about bridge that the probability that each of the four players is dealt exactly one ace is about \(0.11 .\) This means that (a) in every 100 bridge deals, each player has one ace exactly 11 times. (b) in 1 million bridge deals, the number of deals on which each player has one ace will be exactly 110,000 . (c) in a very large number of bridge deals, the percent of deals on which each player has one ace will be very close to \(11 \%\) (d) in a very large number of bridge deals, the average number of aces in a hand will be very close to \(0.11 .\) (e) If each player gets an ace in only 2 of the first 50 deals, then each player should get an ace in more than \(11 \%\) of the next 50 deals.

Spinning a quarter With your forefinger, hold a new quarter (with a state featured on the reverse) upright, on its edge, on a hard surface. Then flick it with your other forefinger so that it spins for some time before it falls and comes to rest. Spin the coin a total of 25 times, and record the results. (a) What's your estimate for the probability of heads? Why? (b) Explain how you could get an even better estimate.

A basketball player claims to make \(47 \%\) of her shots from the field. We want to simulate the player taking sets of 10 shots, assuming that her claim is true. To simulate the number of makes in 10 shot attempts, you would perform the simulation as follows: (a) Use 10 random one-digit numbers, where \(0-4\) are a make and \(5-9\) are a miss. (b) Use 10 random two-digit numbers, where \(00-46\) are a make and \(47-99\) are a miss. (c) Use 10 random two-digit numbers, where \(00-47\) are a make and \(48-99\) are a miss. (d) Use 47 random one-digit numbers, where 0 is a make and \(1-9\) are a miss. (e) Use 47 random two-digit numbers, where \(00-46\) are a make and \(47-99\) are a miss.
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