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Chapter 6: Problem 89

A student has a box containing 25 computer disks, of which 15 are blank and 10 are not. She randomly selects disks one by one and examines each one, terminating the process only when she finds a blank disk. What is the probability that she must examine at least two disks? (Hint: What must be true of the first disk?)

Short Answer

Expert verified
The probability that the student must examine at least two disks to find a blank one is 0.25 or 25%.

Step by step solution

01

Identify the probabilities

Firstly, the student chooses a disk from the box which contains 25 disks, out of which 10 are not blank. Hence, the probability to pick a non-blank disk first is given by the ratio of the number of non-blank disks to the total number of disks. Thus, \( P_1 = \frac{10}{25} = 0.4 \).
02

Conditional Probability

Secondly, as one disk has already been picked and it was not blank, the conditions change for the next selection. Now there are 24 disks remained, out of which 15 are blank. Therefore, the student's successful attempt in the next picking, or finding a blank disk on the second pick, is given by \( P_2 = \frac{15}{24} = 0.625 \).
03

Compute the final probability

The two events described are independent; the result of the first pick does not affect the second one. As the student must examine at least two disks, this means that the first disk she picks up needs to be a non-blank one and the second one should be blank. Therefore, final probability will be the product of \( P_1 \) and \( P_2 \). Consequently, the final probability \( P = P_1 * P_2 = 0.4 * 0.625 = 0.25 \).

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Key Concepts

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

Conditional Probability
Imagine you are picking items from a box one by one. Conditional probability comes into play when the outcome of these picks depends on the results of the previous ones. When the student in the exercise selects a non-blank disk first, it changes the conditions of what's left in the box for the next pick. This is because the total count of non-blank disks decreases.
The key point here is understanding that conditional probability describes the likelihood of an event occurring given that another event has already happened.
For instance, after picking a non-blank disk first, we're interested in finding a blank disk in the second pick with fewer disks available. The probability of this happening is recalculated based on the new conditions, leading to the equation: \[P_2 = \frac{15}{24} = 0.625\] This fraction represents the likelihood of selecting a blank disk after a non-blank has been removed.
Independent Events
In probability theory, independent events are those whose outcomes do not affect each other. However, in the exercise, there's a mix of concepts.
The choice of the first disk does alter conditions for the second pick due to the reduced number of disks, which poses a conditional scenario, yet these can be calculated independently for the purpose of the exercise because each disk pick is initially random.
Here, the step-by-step solution considers the independence in terms of calculating the probability of two different necessary outcomes for the main event sequence (finding non-blank and then blank).
  • The student picks a non-blank disk first (Outcome does not restrict second pick unless in counts).
  • Then picks a blank disk next (Calculated from changed conditions).
This amalgamation helps students comprehend isolated probability effects before connecting outcomes.
Probability Calculation
Probability calculation in this context involves taking a step-by-step approach to determine the likelihood of multiple dependent or independent events occurring in a sequence. First, each event is examined by itself, and then combined to find the overall probability for the sequence.
In our example exercise, first, calculate the probability of picking a non-blank disk: \[P_1 = \frac{10}{25} = 0.4\] After this disk is picked, new probabilities need to be calculated due to fewer available disks. For the blank disk in the next pick, re-calculate:\[P_2 = \frac{15}{24} = 0.625\] The final step is combining these probabilities because you need to know the chance of both happening consecutively, using the rule for combining probabilities of independent events:\[P = P_1 \times P_2 = 0.4 \times 0.625 = 0.25\] This shows the student the systematic way to find the probability of needing to pick at least two disks.

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Most popular questions from this chapter

Many cities regulate the number of taxi licenses, and there is a great deal of competition for both new and existing licenses. Suppose that a city has decided to sell 10 new licenses for \(\$ 25,000\) each. A lottery will be held to determine who gets the licenses, and no one may request more than three licenses. Twenty individuals and taxi companies have entered the lottery. Six of the 20 entries are requests for 3 licenses, 9 are requests for 2 licenses, and the rest are requests for a single license. The city will select requests at random, filling as many of the requests as possible. For example, the city might fill requests for \(2,3,1\), and 3 licenses and then select a request for \(3 .\) Because there is only one license left, the last request selected would receive a license, but only one. a. An individual has put in a request for a single license. Use simulation to approximate the probability that the request will be granted. Perform at least 20 simulated lotteries (more is better!). b. Do you think that this is a fair way of distributing licenses? Can you propose an alternative procedure for distribution?

Of the 60 movies reviewed last year by two critics on their joint television show, Critic 1 gave a "thumbs-up" rating to 15 , Critic 2 gave this rating to 20 , and 10 of the movies were rated thumbs-up by both critics. Suppose that 1 of these 60 movies is randomly selected. a. Given that the movie was rated thumbs-up by Critic 1 , what is the probability that it also received this rating from Critic \(2 ?\) b. If the movie did not receive a thumbs-up rating from Critic 2, what is the probability that it also did not receive a thumbs up rating from Critic \(1 ?\) (Hint: Construct a table with two rows for the first critic [for "up" and "not up"] and two columns for the second critic: then enter the relevant probabilities.)

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