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There are three urns \(A, B\) and \(C\). Urn \(A\) contains 4 white balls and 5 blue balls. \(\operatorname{Urn} B\) contains 4 white balls and 3 blue balls. \(\operatorname{Um} C\) contains 2 white balls and 4 blue balls. One ball is drawn from each of these urns. What is the probability that out of these three balls drawn, two are white balls and one is a blue ball. [CBSE-96]

Short Answer

Expert verified
The probability is \(\frac{16}{47}\).

Step by step solution

01

Determine Total Outcomes

Each urn has balls drawn from them, thus for each urn we have to determine the number of total possible outcomes. From urn A, there are \(4 + 5 = 9\) balls. From urn B, there are \(4 + 3 = 7\) balls. From urn C, there are \(2 + 4 = 6\) balls. Therefore, the total possible outcomes when drawing one ball from each urn is \(9 \times 7 \times 6 = 378\).
02

Consider Favorable Outcomes for 2 White, 1 Blue

We need to find the scenarios where two are white balls and one is blue. The possible cases are: 1. Case 1: White from urn A, white from urn B, and blue from urn C. 2. Case 2: White from urn A, blue from urn B, and white from urn C. 3. Case 3: Blue from urn A, white from urn B, and white from urn C.
03

Calculate Favorable Outcomes for Each Case

For each case:**Case 1:** Probability of white from A is \(\frac{4}{9}\), white from B is \(\frac{4}{7}\), blue from C is \(\frac{4}{6}\). Hence, probability for this is \(\frac{4}{9} \times \frac{4}{7} \times \frac{4}{6} = \frac{64}{378}\).**Case 2:** Probability of white from A is \(\frac{4}{9}\), blue from B is \(\frac{3}{7}\), white from C is \(\frac{2}{6}\). Hence, probability for this is \(\frac{4}{9} \times \frac{3}{7} \times \frac{2}{6} = \frac{24}{378}\).**Case 3:** Probability of blue from A is \(\frac{5}{9}\), white from B is \(\frac{4}{7}\), white from C is \(\frac{2}{6}\). Hence, probability for this is \(\frac{5}{9} \times \frac{4}{7} \times \frac{2}{6} = \frac{40}{378}\).
04

Sum the Probabilities of All Cases

Finally, to get the total probability for getting two whites and one blue ball in any order, sum up the probabilities from each case:\[\frac{64}{378} + \frac{24}{378} + \frac{40}{378} = \frac{128}{378} = \frac{32}{94} = \frac{16}{47}\]Therefore, the probability is \(\frac{16}{47}\).

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Key Concepts

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

Combinatorics
Combinatorics is the field in mathematics that deals with counting, arranging, and analysing the ways in which objects can be selected and organized. It is a fundamental skill in probability as it helps in determining the total possible outcomes of a scenario. In the urn problem, combinatorics is employed to calculate the total number of ways a ball can be drawn from each of three urns.

Each urn has a certain number of balls. To find total possible combinations of drawing one ball from each urn, you multiply the number of balls in Urn A, B, and C.
  • Urn A has 9 balls (4 white and 5 blue)
  • Urn B has 7 balls (4 white and 3 blue)
  • Urn C has 6 balls (2 white and 4 blue)
Thus, the total number of outcomes is calculated as:\[9 \times 7 \times 6 = 378\]Understanding these basic principles of combinatorics can simplify the solving of more complex probability problems.
Conditional Probability
Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. In many probability problems, you have to consider each event in the context of other events.

In this problem, we analyze the probabilities of drawing specific types of balls under certain conditions. For instance, if we want one blue and two white balls from the urns, different conditions arise based on which urn you pick a certain colored ball from.
  • Case 1: White ball from A and B, Blue from C
  • Case 2: White ball from A, Blue from B, White from C
  • Case 3: Blue ball from A, White from B and C
Each case reflects a specific chain of events, showing how conditions influence probabilities, such as picking a white ball or a blue ball based on what's available in each urn. The conditional probabilities are applied to each scenario:\[\text{Probability for Case 1} = \frac{4}{9} \times \frac{4}{7} \times \frac{4}{6} = \frac{64}{378}\]Analyzing each condition helps in breaking down the problem into manageable parts, leading to a smoother calculation of overall probabilities.
Problem-Solving Techniques
Effective problem-solving in probability often involves systematic approaches to break down complex problems into simpler parts. In our urn example, strategic problem-solving techniques include dividing the problem into clear cases and calculating probabilities for each separately.

First, you identify all possible cases by organizing them based on specific outcomes. This entails recognizing that the desired result — two white balls and one blue ball — can occur in multiple distinct orders.

Second, calculate the probabilities for each case by identifying how each sequence can occur, like finding the probability of white coming from one urn, blue from another, and so on.
  • Calculate probability for each case separately.
  • Sum the probabilities of all favorable cases.
  • Ensure to simplify fractions to find simplest form of the probability.
By concentrating on each scenario individually and ensuring each step is thoroughly calculated, you can reach a solution that is as simple as \(\frac{16}{47}\) in this particular problem. Problem-solving in probability is about being organized, patient, and methodical.

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Most popular questions from this chapter

The probability for a randomly chosen month to have its 10 th day as Sunday is (a) \(1 / 84\) (b) \(10 / 12\) (c) \(10 / 84\) (d) \(1 / 7\) (e) \(1 / 12\)

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A ten digit number is formed using the digits from zero to nine, every digit being used exactly once. The probability that the number is divisible by four is [Roorkee-1991](a) \(20 / 81\) (b) \(18 / 20\) (c) \(81 / 20\) (d) \(20 / 18\) Solution $$ n=\text { Total number of ways }=10 !-9 ! $$ To find the favourable number of ways, we observe that a number is divisible by 4 if the last two digits are divisible by \(4 .\) Hence, the last two digits can be \(20,40,60,80,12\), \(32,52,72,92,04,24,64,84,16,36,56,76\) \(96,08,28,48,68\) corresponding to each of \(20,40,60,80,04,08\). The remaining 8 places can be filled up in \(8 !\) ways so that the number of ways in this case \(=6.8 !\) And corresponding to remaining 16 possibilities the number of ways \(=16(8 !-7 !)\) Hence \(m=\) favourable number of ways $$ \begin{aligned} &=22.8 !-16.7 ! \\ &\therefore \text { The required probability }=m / n \\ &=\frac{22.8 !-16.7 !}{10 !-9 !} \\ &=\frac{22.8-16}{109.8-9.8}=\frac{160}{648}=\frac{20}{81} \end{aligned} $$

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