Question

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. Suppose also that the hosts randomly select two answers from those submitted with replacement. a. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

Short answer

a. The probability that both selected answers are correct with replacement is 0.00390625. b. The probability that both selected answers are correct without replacement is 0.00383356. The probability of selecting correct answers is slightly higher when selection is done with replacement.

Step-by-step solution

Calculate Probability With Replacement

Firstly, evaluate the probability when answers are selected with replacement. The probability of selecting a correct answer is the number of correct answers divided by the total number of answers. In this case, \( \frac{50}{800} = 0.0625 \). Since the selection is with replacement, the chances remain the same for the second selection too. Hence, the probability that both selected answers are correct is \( 0.0625 * 0.0625 = 0.00390625 \).

Calculate Probability Without Replacement

Next, evaluate the probability when answers are selected without replacement. The probability of selecting a correct answer in the first draw remains the same, i.e. \( \frac{50}{800} = 0.0625 \). For the second draw, the number of correct answers left is 49 and the total number of answers left is 799. Therefore, the conditional probability of drawing a correct answer in the second draw given a correct answer on the first draw is \( \frac{49}{799} \approx 0.061327 \). Hence, the probability that both selected answers are correct is \( 0.0625 * 0.061327 = 0.00383356 \)

Compare the Probabilities

Comparing the probabilities from step 1 and step 2, we find that the probability that both the answers are correct is higher when answers are selected with replacement. This is because with replacement, the set of answers remains unchanged for the 2nd draw, offering more chances to draw a correct answer, while without replacement, the set of answers is reduced for the 2nd draw.

Key concepts

Conditional Probability

Conditional probability is a fundamental concept in statistics that deals with the likelihood of an event occurring, given that another event has already occurred. It's represented mathematically as the probability of event A happening given that B has already happened, written as P(A|B). In the exercise provided, we explore conditional probability by calculating the chance that two correct answers are selected without replacement. This introduces a dependence between selections; the outcome of the first draw affects the probability of the second. After one correct answer is selected, there is one fewer correct answer available, impacting the odds for the second draw. Understanding this dynamic is crucial in many real-world applications, such as predictive modeling or strategy games where events are interlinked.

So, how do you calculate it? Well, if the probability of selecting the first correct answer is P(A), and the probability of selecting another correct answer after the first correct answer has been drawn is P(B|A), the combined probability of both events A and B occurring is P(A) * P(B|A). In the exercise, this gave us a probability of \( 0.0625 * 0.061327 = 0.00383356 \) after we computed each step separately. The decrease in total number of answers and number of correct answers after the first selection without replacement drives the fundamental difference from with-replacement scenarios.

Probability Calculation

Probability calculation involves determining the chance that a given event will happen. It can range from very basic scenarios, like flipping a coin, to far more complex cases, like the one presented in our puzzle involving random selections with and without replacement. The calculations often require a division of the number of desired outcomes by the total number of possible outcomes to find a probability value between 0 (impossible event) and 1 (certain event).

Looking at the provided exercise, the first step with replacement simply required us to square the initial probability of selecting one correct answer, since each selection is independent and has the same total number of outcomes (\( 0.0625 * 0.0625 = 0.00390625 \) ). Without replacement, the process was a bit more involved because the total number of outcomes decreases with each selection. The ability to carry out such calculations deftly is essential for not just solving textbook problems but for making informed decisions in various spheres of life, including finance, health, and science.

Random Selection

Random selection is the process of selecting items from a population or set where each item has an equal chance of being chosen. It is a fundamental part of probability and statistics and ensures the fairness of the selection process, which is essential in drawing valid conclusions from a sample. In the context of our exercise, this concept is illustrated through the action of randomly picking out answers from a larger pool.

When we select with replacement, as in part (a) of the exercise, the selected item is put back into the population, and thus the composition of the population remains unchanged for the next selection. However, when we select without replacement, as in part (b), the second selection is made from a modified population—one answer fewer in case of a correct pick. Such a method is used in various scenarios, from lottery drawings to scientific sampling, and affects the probability outcomes of subsequent draws. Understanding the mechanics of both methods is crucial for anyone tackling problems in probability theory and real-world situations where the difference has significant implications.