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The probability that a new car battery functions for over 10,000 miles is .8, the probability that it functions for over 20,000 miles is \(.4\), and the probability that it functions for over 30,000 miles is .1. If a new car battery is still working after 10,000 miles, what is the probability that (a) its total life will exceed 20,000 miles; (b) its additional life will exceed 20,000 miles?

Short Answer

Expert verified
The short answer is: a) The probability that the total life of a new car battery will exceed 20,000 miles, given that it is still working after 10,000 miles, is \(0.5\). b) The probability that the additional life of a new car battery will exceed 20,000 miles, given that it is still working after 10,000 miles, is \(0.125\).

Step by step solution

01

Calculate the probability for part (a)

We know that, \(P(A)\) = 0.8 \(P(B)\) = 0.4 To find the probability that the total life of the car battery will exceed 20,000 miles (which is event B) given that it is still working after 10,000 miles (which is event A), we will use the conditional probability formula: \(P(B|A) = \frac{P(A \cap B)}{P(A)}\) Note that in this case, event B implies event A, meaning that if the car battery functions for over 20,000 miles, then it must have functioned for more than 10,000 miles. So, \(P(A \cap B) = P(B) = 0.4\). Now, we can find the conditional probability: \(P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.4}{0.8} = 0.5\)
02

Calculate the probability for part (b)

We can think of additional life exceeding 20,000 miles as a car battery functioning for more than 30,000 miles in total. We know that, \(P(C)\) = 0.1 So, the probability that the additional life of the car battery will exceed 20,000 miles (which is event C) given that it is still working after 10,000 miles (which is event A) can be found using the conditional probability formula: \(P(C|A) = \frac{P(A \cap C)}{P(A)}\) Again, event C implies event A, so \(P(A \cap C) = P(C) = 0.1\). Now, we can find the conditional probability: \(P(C|A) = \frac{P(A \cap C)}{P(A)} = \frac{0.1}{0.8} = \frac{1}{8} = 0.125\)
03

Present the final probabilities for (a) and (b)

a) The probability that the total life of the car battery will exceed 20,000 miles, given that it is still working after 10,000 miles is 0.5. b) The probability that the additional life of the car battery will exceed 20,000 miles, given that it is still working after 10,000 miles is 0.125.

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random events and quantifying the likelihood of various outcomes. It is based on the concept that all possible outcomes of an event can be listed, and the probability of each outcome can be determined. In practical terms, probability can range from 0 to 1, where 0 means an event is impossible and 1 means it is certain to occur.

For example, when trying to determine the likelihood that a new car battery will function past certain mileage benchmarks, we are dealing with a sequence of events, each having its own probability. Probability theory not only helps us find the likelihood of a single event occurring, such as functioning for over 10,000 miles, but also allows us to analyze more complex situations, like the likelihood of a battery functioning for over 20,000 miles given that it has already surpassed 10,000 miles.
Bayes' Theorem
Bayes' theorem is a powerful formula used in probability theory to calculate the probability of an event based on prior knowledge of conditions that might be related to the event. It is named after the Reverend Thomas Bayes and can be seen as an extension of conditional probabilities.

The theorem can be formalized as:
\[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\],
where \(P(A|B)\) is the probability of event \(A\) occurring given that \(B\) has occurred, \(P(B|A)\) is the probability of \(B\) given \(A\), \(P(A)\) is the probability of \(A\) and \(P(B)\) is the probability of \(B\). What makes Bayes' theorem so powerful is its ability to update predictions based on new evidence, a fundamental tool in statistical inference and various applications like spam filtering and medical diagnosis.
Probability Formulas
Probability formulas are essential tools to quantify the likelihood of various events. Some basic formulas include the rule of addition for finding the probability of the union of two events, and the multiplication rule for finding the probability of the intersection of two events. For disjoint events, the rule of addition is simply \(P(A \cup B) = P(A) + P(B)\).

However, when events can occur simultaneously, their intersection probability needs to be subtracted to avoid double-counting: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). The multiplication rule states that for two independent events, the probability of both occurring is \(P(A \cap B) = P(A) \cdot P(B)\). For conditional probabilities, we use: \(P(A|B) = \frac{P(A \cap B)}{P(A)}\), which tells us the likelihood of event \(A\) given that \(B\) has occurred.
Event Intersection Probability
Event intersection probability refers to the likelihood of two events occurring simultaneously. In the context of our exercise, when we talk about a car battery functioning past 10,000 miles (event A) and also functioning beyond 20,000 miles (event B), we're interested in the intersection of these two events, represented as \(P(A \cap B)\).

When two events must happen in succession for the second condition to be met (like the battery surviving first 10,000 miles before reaching 20,000), the intersection probability is equivalent to the probability of the second event: \(P(A \cap B) = P(B)\), since event B cannot happen without event A. This concept is crucial for understanding conditional probability and is applied in our example to determine the probability that a car battery continues to function after already reaching a certain milestone.

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Most popular questions from this chapter

The Ballot Problem. In an election, candidate \(A\) receives \(n\) votes and candidate \(B\) receives \(m\) votes, where \(n>m\). Assuming that all of the \((n+m) ! / n ! m !\) orderings of the votes are equally likely, let \(P_{n, m}\) denote the probability that \(A\) is always ahead in the counting of the votes. (a) Compute \(P_{2,1}, P_{3,1}, P_{3,2}, P_{4,1}, P_{4,2}, P_{4,3}\). (b) Find \(P_{n, 1}, P_{n, 2}\) (c) Based on your results in parts (a) and (b), conjecture the value of \(P_{n, m}\). (d) Derive a recursion for \(P_{n, m}\) in terms of \(P_{n-1, m}\) and \(P_{n, m-1}\) by conditioning on who receives the last vote. (e) Use part (d) to verify your conjecture in part (c) by an induction proof on \(n+m\).

Consider the gambler's ruin problem with the exception that \(A\) and \(B\) agree to play no more than \(n\) games. Let \(P_{n, i}\) denote the probability that \(A\) winds up with all the money when \(A\) starts with \(i\) and \(B\) with \(N-i\). Derive an equation for \(P_{n, i}\) in terms of \(P_{n-1, i+1}\) and \(P_{n-1, i-1}\) and compute \(P_{7,3}\), \(N=5\)

Consider two independent tosses of a fair coin. Let \(A\) be the event that the first toss lands heads, let \(B\) be the event that the second toss lands heads, and let \(C\) be the event that both land on the same side. Show that the events \(A\), \(B, C\) are pairwise independent-that is, \(A\) and \(B\) are independent, \(A\) and \(C\) are independent, and \(B\) and \(C\) are independent-but not independent.

Independent trials that result in a success with probability \(p\) are successively performed until a total of \(r\) successes is obtained. Show that the probability that exactly \(n\) trials are required is $$ \left(\begin{array}{l} n-1 \\ r-1 \end{array}\right) p^{r}(1-p)^{n-r} $$ Use this result to solve the problem of the points (Example 4i). HINT. In order for it to take \(n\) trials to obtain \(r\) successes, how many successes must occur in the first \(n-1\) trials?

If two fair dice are rolled, what is the conditional probability that the first one lands on 6 given that the sum of the dice is \(i\) ? Compute for all values of \(i\) between 2 and 12 .

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