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An urn contains 4 white and 4 black balls. We randomly choose 4 balls. If 2 of them are white and 2 are black, we stop. If not, we replace the balls in the urn and again randomly select 4 balls. This continues until exactly 2 of the 4 chosen are white. What is the probability that we shall make exactly \(n\) selections?

Short Answer

Expert verified
The probability of making exactly n selections until we get 2 white and 2 black balls in a single draw for the first time is given by the expression: \[P(\text{exactly } n \text{ selections}) = \left(\frac{17}{35}\right)^{n-1} \cdot \frac{18}{35}\]

Step by step solution

01

Calculate the probability of getting 2 white and 2 black balls in one draw

There are \(\binom{8}{4}\) ways to choose 4 balls out of 8 without regard to the color. We can have 2 white and 2 black balls in one draw, which can be chosen in \(\binom{4}{2}\binom{4}{2}\) ways. Therefore, the probability of getting 2 white and 2 black balls in one draw is: \[P = \frac{\binom{4}{2}\binom{4}{2}}{\binom{8}{4}}\]
02

Calculate the probability of not getting 2 white and 2 black balls in one draw

The complementary event is failing to get 2 white and 2 black balls. We can calculate the probability of this complementary event by subtracting the probability of success from 1: \[P(\text{not getting } 2 \text{ white and } 2 \text{ black balls}) = 1-P\]
03

Evaluate the probabilities

Now, we calculate these probabilities: \[P = \frac{\binom{4}{2}\binom{4}{2}}{\binom{8}{4}} = \frac{6 \cdot 6}{70} = \frac{36}{70} = \frac{18}{35}\] \[P(\text{not getting } 2 \text{ white and } 2 \text{ black balls}) = 1 - \frac{18}{35} = \frac{17}{35}\]
04

Calculate the probability of making exactly n selections

To find the probability of making exactly n selections until we get 2 white and 2 black balls in a single draw for the first time, we have to fail in the first n-1 draws and succeed in the nth draw. Since these events are independent, we can multiply their probabilities: \[P(\text{exactly } n \text{ selections}) = \left(\frac{17}{35}\right)^{n-1} \cdot \frac{18}{35}\] The final answer is given by the expression: \[P(\text{exactly } n \text{ selections}) = \left(\frac{17}{35}\right)^{n-1} \cdot \frac{18}{35}\]

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Key Concepts

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

Combinatorics
Combinatorics is the branch of mathematics that deals with counting, arranging, and grouping items. In the context of this exercise, we are interested in calculating the number of ways in which we can select 4 balls from a total of 8 balls in the urn, which includes both black and white balls.
When we talk about combinations, we use the binomial coefficient notation, represented as \( \binom{n}{k} \), which means choosing \( k \) items from \( n \) total items without considering the order. In this exercise, to choose 4 balls out of 8, we use \( \binom{8}{4} \) since the order of selection doesn't matter.

Specifically for this problem, we not only want to choose 4 balls but also need a certain configuration: 2 white and 2 black. The number of ways to choose 2 white balls out of 4 white ones is \( \binom{4}{2} \), and similarly, the number of ways to choose 2 black balls from 4 black ones is again \( \binom{4}{2} \). Hence, the total number of desirable selections can be calculated as the product of the two combinations: \( \binom{4}{2}\binom{4}{2} \).
The power of combinatorics lies in allowing us to systematically count various configurations even in complex scenarios.
Random Selection
Random selection involves picking items from a set without any specific pattern or predictability. In these problems, we are dealing with probability and chance. The choice of balls from an urn is a classic representation of random selection in probability theory.
Every time we select balls from the urn, each ball has an equal probability of being chosen. This randomness is key to calculating the probability of a certain event occurring, such as drawing exactly 2 white and 2 black balls. The unpredictability of the outcome means that each draw has to consider all possible combinations of selected balls.

To make sure that our selections reflect true randomness, we assume that each ball is equally likely to be picked. This means when a problem states that the draw is random, it confirms no bias or pre-conditions influence the selection process, highlighting the core idea of uniform probability distribution. Ensuring a random selection helps us apply our knowledge of probability accurately to predict outcomes.
Complementary Probability
Complementary probability is a useful concept that involves determining the probability that a certain event does not occur. This principle is often simpler, especially if calculating the direct probability is complex.
For any event \( A \), the probability of the event not happening is given by \( 1 - P(A) \). Here in the urn problem, the event \( A \) is drawing exactly 2 white and 2 black balls. If we know the probability of this happening, \( P(A) \), the complementary probability of not drawing 2 white and 2 black balls is \( 1 - P(A) \).

Understanding this concept allows us to efficiently determine the probability of continuous failure until success. In this exercise, to find the probability of exactly \( n \) selections, recognizing that it entails failing \( n-1 \) times and succeeding on the \( n \)th time, is key. Using complementary probability makes it easy to first calculate all possible successful outcomes and then use subtraction to find the failure rate, streamlining complex probability problems.

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Most popular questions from this chapter

At time \(0,\) a coin that comes up heads with probability \(p\) is flipped and falls to the ground. Suppose it lands on heads. At times chosen according to a Poisson process with rate \(\lambda,\) the coin is picked up and flipped. (Between these times the coin remains on the ground.) What is the probability that the coin is on its head side at time \(t ?\) Hint What would be the conditional probability if there were no additional flips by time \(t,\) and what would it be if there were additional flips by time \(t ?\)

There are \(k\) types of coupons. Independently of the types of previously collected coupons, each new coupon collected is of type \(i\) with probability \(p_{i}, \sum_{i=1}^{k} p_{i}=1 .\) If \(n\) coupons are collected, find the expected number of distinct types that appear in this set. (That is, find the expected number of types of coupons that appear at least once in the set of \(n\) coupons.)

When coin 1 is flipped, it lands on heads with probability \(.4 ;\) when coin 2 is flipped, it lands on heads with probability .7. One of these coins is randomly chosen and flipped 10 times. (a) What is the probability that the coin lands on heads on exactly 7 of the 10 flips? (b) Given that the first of these ten flips lands heads, what is the conditional probability that exactly 7 of the 10 flips land on heads?

Suppose that two teams play a series of games that ends when one of them has won \(i\) games. Suppose that each game played is, independently, won by team \(A\) with probability \(p .\) Find the expected number of games that are played when (a) \(i=2\) and (b) \(i=3 .\) Also, show in both cases that this number is maximized when \(p=\frac{1}{2}\)

Each of 500 soldiers in an army company independently has a certain disease with probability \(1 / 10^{3}\). This disease will show up in a blood test, and to facilitate matters, blood samples from all 500 soldiers are pooled and tested. (a) What is the (approximate) probability that the blood test will be positive (that is, at least one person has the disease)? Suppose now that the blood test yields a positive result. (b) What is the probability, under this circumstance, that more than one person has the disease? One of the 500 people is Jones, who knows that he has the disease. (c) What does Jones think is the probability that more than one person has the disease? Because the pooled test was positive, the authorities have decided to test each individual separately. The first \(i-1\) of these tests were negative, and the \(i\) th one-which was on Jones-was positive. (d) Given the preceding, scenario, what is the probability, as a function of \(i,\) that any of the remaining people have the disease?

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