/*! 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 10 A bag \(x\) contains 3 white bal... [FREE SOLUTION] | 91Ó°ÊÓ

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A bag \(x\) contains 3 white balls and 2 black balls and another bag \(y\) contains 2 white balls and 4 black balls. A bag and a ball out of it are picked at random. The probability that the ball is white is (a) \(3 / 5\) (b) \(7 / 15\) (c) \(1 / 2\) (d) None

Short Answer

Expert verified
The probability that the ball is white is \( \frac{7}{15} \) (option b).

Step by step solution

01

Understand the Problem

We have two bags, each with a different number of white and black balls. We need to find the probability that a randomly selected ball from a randomly selected bag is white.
02

Determine the Probability of Choosing Each Bag

Since there are two bags, and each bag is equally likely to be chosen, the probability of choosing either bag is \( \frac{1}{2} \).
03

Calculate the Probability of Picking a White Ball from Each Bag

- For bag \(x\), which has 3 white and 2 black balls, the probability of picking a white ball is \( \frac{3}{5} \).- For bag \(y\), which has 2 white and 4 black balls, the probability of picking a white ball is \( \frac{2}{6} = \frac{1}{3} \).
04

Calculate the Total Probability of Picking a White Ball

Using the law of total probability:- Probability from bag \(x\): \( \frac{1}{2} \times \frac{3}{5} = \frac{3}{10} \).- Probability from bag \(y\): \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \).Add these probabilities together to get the total probability: \( \frac{3}{10} + \frac{1}{6} = \frac{9}{30} + \frac{5}{30} = \frac{14}{30} = \frac{7}{15} \).

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Key Concepts

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

Law of Total Probability
The law of total probability is a fundamental concept in probability theory that allows us to find the probability of an event by considering all possible ways that event can occur. In our exercise with the bags, the event we're interested in is picking a white ball. There are two pathways for this to happen: picking a white ball from bag \(x\) or picking a white ball from bag \(y\). Each pathway has its own probability.To apply the law of total probability, we start by calculating the probability of each smaller event (in this case, selecting a specific bag and then picking a white ball). Since each bag is equally likely to be chosen, the probability of selecting either bag is \(\frac{1}{2}\). Then, we multiply this by the probability of selecting a white ball from each bag:
  • For bag \(x\): the chance is \(\frac{3}{5}\), leading to a combined probability of \(\frac{1}{2} \times \frac{3}{5} = \frac{3}{10}\).
  • For bag \(y\): the chance is \(\frac{1}{3}\), leading to a combined probability of \(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\).
Finally, we add these probabilities together to get the total probability of picking a white ball, which is \(\frac{7}{15}\). This method ensures that every potential path to our desired outcome is considered.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. It’s a key concept for understanding how probabilities change based on new information. In our problem, determining the probability of picking a white ball given we have already chosen a specific bag exemplifies conditional probability.For bag \(x\), the probability of picking a white ball once this bag is selected is seen as \(\frac{3}{5}\). This is because we’re only considering outcomes within bag \(x\), and the chances of drawing a white ball given you already picked bag \(x\) shifts from a global perspective to a local one.Similarly, for bag \(y\), this conditional probability is calculated to be \(\frac{1}{3}\). These probabilities illustrate how conditional probabilities adjust your likelihood of an event once some conditions are set (like choosing a bag first).Understanding how conditions affect the probability of events can be crucial in both day-to-day reasoning and complex statistical tasks. It represents a focus on specifics rather than the entire sample space.
Combinatorics
Combinatorics is the branch of mathematics dealing with the counting, arrangement, and combination of objects. Although the given problem does not explicitly call for complex combinatorial calculations, its principles underpin why we can logically break the problem into more manageable parts. Choosing a ball from one of the bags involves basic combinatorial ideas since we're dealing with combinations of different choices — one of two bags and a specific color of ball per bag. Combinatorics helps us determine the number of ways we can select items under different configurations, which is central when calculating probabilities. In our example, the configuration involves:
  • Selecting one out of two bags.
  • Choosing specific balls (white or black) from a selected bag with a defined combination of balls.
Thus, while direct combinatorial calculation is not utilized here, understanding how these calculations work is critical when dealing with more complex problems or when problems involve greater numbers of choices and selections. By breaking down larger problems, combinatorics simplifies our approach, making tackling probability scenarios much easier.

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Most popular questions from this chapter

\(A\) 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 $$ =p: p^{\prime}=\frac{34}{35}: \frac{13}{28}=\frac{952}{1540}: \frac{715}{1540}=952: 715 $$

Each coefficient in the equation \(a x^{2}+b x+\) \(c=0\) is determined by throwing an ordinary die. The probability that the equation will have equal roots is [Roorkee-1998] (a) \(216 / 5\) (b) \(261 / 5\) (c) \(5 / 261\) (d) \(5 / 216\)Solution $$ \begin{aligned} &\text { Roots equal } \Rightarrow b^{2}-4 a c=0 \\ &\therefore\left(\frac{b}{2}\right)^{2}=a c \end{aligned} $$ Each coefficient is an integer, so we consider the following cases: \(b=1 \quad \therefore \frac{1}{4}=a c\) No integral values of \(a\) and \(c\) \(\begin{array}{ll}b=2 & 1=a c \\ b=3 & 9 / 2=a c\end{array} \quad \therefore(1,1)\) No integral values of \(a\) and \(c\) \(b=4 \quad 4=a c\) \(\therefore(1,4),(2,2),(4,1)\) \(b=5 \quad 25 / 2=a c\), No integral values of \(a\) and \(c\) \(b=6 \quad 9=a c \quad \therefore 3,3\) Thus we have 5 favourable ways for \(b=2,4,6\) Total number of equations is \(6.6 .6=216\) \(\therefore\) Required probability is \(5 / 216\) aragraph for Questions 29 to 31 A fair die is tossed repeatedly until a 6 is obtained. Let \(X\) denote the number of tosses required

For a biased die, the probabilities for different faces to turn up are \(\begin{array}{lcccccc}\text { Face } & : 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Probability: } & 0.2 & 0.22 & 0.11 & 0.25 & 0.05 & 0.17\end{array}\) The die is tossed and you are told that either face 4 or face 5 has turned up. The probability that it is face 4 is (a) \(1 / 6\) (b) \(1 / 4\) (c) \(5 / 6\) (d) None

A letter is known to have come from either TATANAGAR or CALCUTTA. On the envelope, just two consecutive letters. TA are visible. The probability that the letters have come from CALCUTTA is (a) \(1 / 3\) (b) \(4 / 11\) (c) \(5 / 12\) (d) None of these (b) Let \(A\) : 'the event that letters came from TATANAGAR' B: 'the event that letters came from CALCUTTA' \(C\) : 'the event that two consecutive letters visible by \(T A^{\prime}\) Then \(P(A)=1 / 2, P(B)=1 / 2, P(C / B)=1 / 7\) Hence, by Bayes' theorem $$ P(B / C)=\frac{P(B) \times P(C / B)}{P(A) P(C / A)+P(B) P(C / B)}=\frac{4}{11} $$

A random variable \(X\) is specified by the following distribution law: \(\begin{array}{lccc}X: & 2 & 3 & 4 \\ P(X=x): & 0.3 & 0.4 & 0.3\end{array}\) Then the variance of this distribution is (a) \(0.6\) (b) \(0.7\) (c) \(0.77\) (d) \(1.55\) ution $$ \begin{aligned} &\text { (a) } \text { Mean }=(2)(0.3)+(3)(0.4)+(4)(0.3)=3 \\ &x^{2}=\operatorname{Variance}(x) \sum(x-\bar{x})^{2} P \\ &\quad=(2-3)^{2}(0.3)+(3.3)^{2}(0.4)+(4-3)^{2} \times 0.3 \\ &=0.6 \end{aligned} $$

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