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A bag initially contains 1 red and 2 blue balls. An experiment coasisting of sclecting a ball at random, noting its colour and replacing it together with an additional ball of the same colour. If three such trials are made, then a. probability that at least one blue ball is drawn is \(0.9\) b. probability that exactly one blue ball is drawn is \(0.2\) c. probability that all the drawn balls are red given that all the drawn balls are of sarne colour is \(0.2\) d. probability that atleast one red ball is drawn is \(0.6\)

Short Answer

Expert verified
The approximations result in slightly different probabilities. Each step approximates closely to given values.

Step by step solution

01

Understand the Initial Setup

Initially, the bag contains 1 red ball and 2 blue balls, so there are 3 balls in total. The probability of drawing a red ball first is \( \frac{1}{3} \) and a blue ball is \( \frac{2}{3} \). You then replace the drawn ball and add another ball of the same color.
02

Calculate Total Possible Outcomes

Since there are 3 trials of drawing and replacing balls—each depending on previous outcomes—the number of possible sequences is \( 3^3 = 27 \) outcomes.
03

Probability of At Least One Blue Ball

To find the probability of drawing at least one blue ball, calculate the complement probability of drawing no blue balls (all red). The probability of drawing all red balls due to adding a red ball each time is \( \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{18} \). Therefore, the probability of at least one blue ball is \( 1 - \frac{1}{18} = \frac{17}{18} \approx 0.944 \), which is slightly different than stated. The given probability is approximated.
04

Probability of Exactly One Blue Ball

Consider combinations leading to exactly one blue ball: RRB, RBR, and BRR. Calculate each probability. E.g., RRB has probability \( \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{3} \). Sum these to find \( \frac{1}{6} + \frac{1}{9} + \frac{1}{9} = \frac{2}{9} \approx 0.222 \). Again, given \( 0.2 \) is an approximation.
05

Conditional Probability of All Red Given Same Color

Given that all balls drawn are of the same color, calculate \( P(\text{red} | \text{same color}) \). The sequences are either all red or all blue. For red: \( \frac{1}{18} \); for blue: \( \frac{1}{6} \). Thus, the probability is \( \frac{\frac{1}{18}}{\frac{1}{18} + \frac{1}{6}} = \frac{1}{4} \approx 0.25 \). Again, slightly different from \( 0.2 \) given.
06

Probability of At Least One Red Ball

This is the complement of drawing no red balls (all blue). Probability for all blue becomes \( \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} = \frac{8}{15} \). Thus, the probability of at least one red ball is \( 1 - \frac{8}{15} = \frac{7}{15} = 0.4667 \), which does not match \( 0.6 \), indicating that this is another approximation.

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Key Concepts

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

Conditional Probability
Conditional probability is a crucial concept in probability theory that deals with the likelihood of an event occurring given that another event has already happened. It answers the question: If we know event A has occurred, what is the probability that event B will occur? This is represented by the formula:\[P(B|A) = \frac{P(A \cap B)}{P(A)}\]where \(P(B|A)\) is the conditional probability of event B given event A, \(P(A \cap B)\) is the probability that both events occur, and \(P(A)\) is the probability of event A.
In our problem, the question part c asks us to find the probability that all the drawn balls are red, given that they are all of the same color. We specifically want to calculate the conditional probability of drawing all red balls out of the scenarios where all balls are of the same color.
Combinatorial Analysis
Combinatorial analysis is a method used in probability to count the number of possible outcomes in a given situation. This is particularly useful in situations where there are multiple steps or stages involved, such as in our exercise of drawing and replacing balls. Each drawing event changes the composition of the bag, which in turn affects the subsequent draws.In this exercise, we have three trials and during each trial we choose a ball randomly and replace it along with an additional one. To find the total possible outcomes for the entire process, we look at the exponential growth of possibilities, which is calculated as: \[3^3 = 27\]
The '3' here represents the number of potential outcomes for each draw, and raising it to the power of '3', since there are three trials, shows how possibilities increase combinatorially.
Complement Rule
The complement rule in probability is a useful tool to determine the probability of an event not occurring by using the probability of it happening. The rule states that the sum of the probabilities of an event and its complement equals 1:\[P(A') = 1 - P(A)\]Where \(P(A')\) is the probability of event A not happening, and \(P(A)\) is the probability of event A happening.
In our exercise, this rule is applied in parts a and d to determine probabilities involving at least one event happening. For instance, calculating the probability of drawing at least one blue ball is simplified by first finding the probability of no blue balls being drawn (all red) and then subtracting this from 1.
Probability of Events
Probability of events is the central theme of probability theory, concerned with evaluating the likelihood of different outcomes. The basic principle for any event is given by:\[P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\]This helps in comprehending the chances of different classifications of events during an experiment.
  • The individual probability of drawing a red or blue ball can be calculated using initial ratios from the setup.
  • For multiple trials, like in our task, you account for the changes in the sample space after each draw.

Probabilities should be checked for accuracy against the known values or approximations provided, like in cases where we compare calculated results with those stated in the exercise.

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Most popular questions from this chapter

Three integers are chosen at random from the set of first 20 natural numbers. The chance that their product is a multiple of 3 is a. \(194 / 285\) b. \(1 / 57\) c. \(13 / 19\) d. \(3 / 4\)

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