/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A bag contains 9 white balls and... [FREE SOLUTION] | 91Ó°ÊÓ

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A bag contains 9 white balls and 5 black balls. Another bag contains 8 white balls and 6 black balls. One ball is transferred from the first bag into the second, and then a ball is drawn from the latter. The probability that it will be a white ball is (a) \(1 / 2\) (b) \(1 / 5\) (c) \(1 / 21\) (d) \(121 / 210\)

Short Answer

Expert verified
The probability that a drawn ball will be white is \( \frac{121}{210} \) (option d).

Step by step solution

01

Determine Initial Probabilities

Calculate the probability of transferring a white ball and a black ball from the first bag, and the consequences for the second bag.- Probability of transferring a white ball: \( \frac{9}{14} \)- Probability of transferring a black ball: \( \frac{5}{14} \)
02

Calculate Probabilities After White Ball Transferred

If a white ball is transferred, the second bag has 9 white and 6 black balls.- Probability of drawing a white ball from the second bag: \( \frac{9}{15} = \frac{3}{5} \)
03

Calculate Probabilities After Black Ball Transferred

If a black ball is transferred, the second bag has 8 white and 7 black balls.- Probability of drawing a white ball from the second bag: \( \frac{8}{15} \)
04

Combine Results Using Total Probability

Calculate the total probability of drawing a white ball from the second bag using the law of total probability.\[ P(\text{White}) = \left( \frac{9}{14} \cdot \frac{3}{5} \right) + \left( \frac{5}{14} \cdot \frac{8}{15} \right) \]
05

Simplify Final Probability

Simplify the expression to determine the final probability:\[P(\text{White}) = \frac{27}{70} + \frac{40}{210} = \frac{81 + 40}{210} = \frac{121}{210}\]

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Key Concepts

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

Total Probability Theorem
The Total Probability Theorem is a fundamental principle in probability theory. It helps us determine the probability of an event based on several paths or scenarios. Imagine you have multiple ways to achieve the same outcome or event. Each path has a certain probability, and the theorem combines these probabilities.
This theorem is crucial when you're dealing with compound events. In our case, drawing a white ball from the second bag after a ball has been transferred from the first one is a compound event.
To apply the Total Probability Theorem, you break down the problem like this:
  • Identify each possible scenario (e.g., transferring a white or black ball).
  • Calculate the probability for each scenario separately.
  • Combine the probabilities of these scenarios to get the total probability of the main event (drawing a white ball).
This step-by-step approach provides clarity and ensures that every possible outcome is considered.
Probability Theory
Probability Theory is the branch of mathematics that deals with the likelihood of different outcomes. It helps us understand and quantify uncertainty.
In probability, the outcome of an event is a numerical value between 0 and 1. A probability of 0 means the event cannot happen, while a probability of 1 means it will surely happen. Anything in between indicates varying levels of possibility.
The art of probability isn't just about numbers; it involves understanding how these numbers represent real-world situations. Like in this problem, where we have specific basic events:
  • Transferring a white ball from the first bag to the second.
  • Transferring a black ball instead.
  • Drawing a white ball from the second bag afterward.
By assigning probabilities to these events, we can predict the likelihood of drawing a white ball. Understanding probability theory makes it easier to solve such problems by breaking them down into manageable parts.
Mathematical Problem Solving
Mathematical Problem Solving is a skill that involves using various mathematical concepts and logic to find solutions to problems. Each problem can often be broken down into smaller, more manageable parts.

In the given ball transfer problem, the steps walkthrough the process of determining which ball is likely to be drawn from the second bag. Mathematical problem solving here involves:
  • Breaking down the problem: Identify what happens when a ball is transferred and how it affects the second bag.
  • Utilizing known formulas: Use the Total Probability Theorem to piece together the result.
  • Step-by-step logic: By taking small, carefully calculated steps, we reach a solution without jumping to conclusions.
By applying these problem-solving techniques, you enhance your mathematical intuition and ability to tackle complex problems efficiently. This structured approach ensures you consider all aspects of a problem and arrive at a sound conclusion. Each step becomes clearer, and errors can be minimized.

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Most popular questions from this chapter

The probability distribution of a random variable \(X\) is given below: \(\begin{array}{llll}X=x_{i} & 2 & 3 & 4 \\ P:\left(X=x_{i}\right) & 1 / 4 & 1 / 8 & 5 / 8\end{array}\) Then its mean is (a) \(27 / 8\) (b) \(5 / 4\) (c) 1 (d) \(4 / 5\)

\(A\) and \(B\) are two independent events. The probability that both \(A\) and \(B\) occur is \(1 / 6\) and the probability that none of them occurs is \(1 / 3\). The minimum value of probability of occurrence of \(A\) is (a) \(1 / 2\) (b) \(1 / 3\) (c) \(1 / 4\) (d) None of these

\(A\) six faced fair die is thrown until 1 comes. The probability that 1 comes in even number of trials is (a) \(5 / 11\) (b) \(6 / 11\) (c) \(5 / 6\) (d) \(1 / 6\)

In a certain town, \(40 \%\) of the people have brown hair, \(25 \%\) have brown eyes and \(15 \%\) have both brown hair and brown eyes. If a person selected at random from the town has brown hair, the probability that he also has brown eyes is (a) \(1 / 5\) (b) \(3 / 8\) (c) \(1 / 3\) (d) \(2 / 3\)

\(A\) has 3 shares in a lottery containing 3 prizes and 9 blanks. \(B\) has two shares in a lottery containing 2 prizes and 6 blanks; Find theratio of \(A\) 's chance of success to \(B\) 's chance of success. (a) \(927: 715\) (b) \(972: 751\) (c) \(925: 715\) (d) \(715: 972\) olution (c) Since \(A\) has 3 shares in a lottery, his chance of success means that he gets at least 1 prize, that is, he gets either 1 prize or 2 prizes or 3 prizes and his chance of failure means that he gets no prize. It is certain that either he succeeds or fails. If \(p\) denotes his chance of success and \(q\) the chance of his failure, then \(p+q=1\) or \(p=1-q\) We now find \(q \times n=\) total number of ways $$ ={ }^{12} C_{3}=\frac{12 \times 11 \times 10}{1 \times 2 \times 3}=220 $$ Since out of 12 tickets in the lottery, he can draw any 3 tickets by virtue of his having 3 shares in the lottery and \(m=\) favourable number of ways $$ ={ }^{9} C_{3}=\frac{9 \times 8 \times 7}{1 \times 2 \times 3}=84 $$ Since he will fail to draw a prize if all the tickets drawn by him are blanks. $$ \therefore q=\frac{m}{n}=\frac{84}{220}=\frac{21}{55} $$ \(\therefore p=A\) 's chance of success \(=1-\frac{21}{55}=\frac{34}{55}\)Similarly B's chance of success $$ \begin{aligned} p^{\prime} &=1-q^{\prime}=1-\frac{{ }^{6} C_{2}}{{ }^{8} C_{2}}=1-\frac{6 \times 5}{8 \times 7} \\ &=1-\frac{15}{28}=\frac{13}{28} \end{aligned} $$ \(\therefore A\) 's chance of success: B's chance of success $$ =p: p^{\prime}=\frac{34}{35}: \frac{13}{28}=\frac{952}{1540}: \frac{715}{1540}=952: 715 $$

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