Question:
In a toy factory, the machines \(A\), \(B\), and \(C\) are used to manufacture \(30\%\), \(40\%\), and \(30\%\) of the output, respectively. The probabilities of toys made by machines \(A\), \(B\), and \(C\) to be defective are respectively \(2\%\), \(3\%\), and \(1\%\). A toy is taken from the factory and is found to be defective. The probability that it was manufactured by machine \(B\) is
In a toy factory, the machines \(A\), \(B\), and \(C\) are used to manufacture \(30\%\), \(40\%\), and \(30\%\) of the output, respectively. The probabilities of toys made by machines \(A\), \(B\), and \(C\) to be defective are respectively \(2\%\), \(3\%\), and \(1\%\). A toy is taken from the factory and is found to be defective. The probability that it was manufactured by machine \(B\) is
Show Hint
In Bayes' theorem problems, multiply each machine's production probability by its defect probability, then divide the required machine's defective contribution by the total defective contribution.
Updated On: Jun 26, 2026
- \(\dfrac{4}{5}\)
- \(\dfrac{2}{9}\)
- \(\dfrac{3}{4}\)
- \(\dfrac{4}{7}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Write the given probabilities.
Let \(D\) denote the event that the selected toy is defective.
The probabilities of selecting toys manufactured by machines \(A\), \(B\), and \(C\) are \[ P(A)=\frac{30}{100}=\frac{3}{10} \] \[ P(B)=\frac{40}{100}=\frac{2}{5} \] \[ P(C)=\frac{30}{100}=\frac{3}{10} \] The defective probabilities are \[ P(D|A)=\frac{2}{100}=\frac{1}{50} \] \[ P(D|B)=\frac{3}{100} \] \[ P(D|C)=\frac{1}{100} \]
Step 2: Apply Bayes' theorem.
We need to find \[ P(B|D) \] By Bayes' theorem, \[ P(B|D)=\frac{P(B)P(D|B)} {P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)} \]
Step 3: Substitute the values.
\[ P(B|D) = \frac{\frac{2}{5}\cdot\frac{3}{100}} {\frac{3}{10}\cdot\frac{2}{100} +\frac{2}{5}\cdot\frac{3}{100} +\frac{3}{10}\cdot\frac{1}{100}} \] Now, \[ \frac{2}{5}\cdot\frac{3}{100} = \frac{6}{500} = \frac{3}{250} \] Also, \[ \frac{3}{10}\cdot\frac{2}{100} = \frac{6}{1000} \] \[ \frac{2}{5}\cdot\frac{3}{100} = \frac{12}{1000} \] \[ \frac{3}{10}\cdot\frac{1}{100} = \frac{3}{1000} \] Therefore, the denominator is \[ \frac{6}{1000}+\frac{12}{1000}+\frac{3}{1000} = \frac{21}{1000} \]
Step 4: Simplify the probability.
\[ P(B|D) = \frac{\frac{12}{1000}}{\frac{21}{1000}} \] \[ =\frac{12}{21} \] \[ =\frac{4}{7} \]
Step 5: Final conclusion.
Therefore, the probability that the defective toy was manufactured by machine \(B\) is \[ \boxed{\frac{4}{7}} \]
Let \(D\) denote the event that the selected toy is defective.
The probabilities of selecting toys manufactured by machines \(A\), \(B\), and \(C\) are \[ P(A)=\frac{30}{100}=\frac{3}{10} \] \[ P(B)=\frac{40}{100}=\frac{2}{5} \] \[ P(C)=\frac{30}{100}=\frac{3}{10} \] The defective probabilities are \[ P(D|A)=\frac{2}{100}=\frac{1}{50} \] \[ P(D|B)=\frac{3}{100} \] \[ P(D|C)=\frac{1}{100} \]
Step 2: Apply Bayes' theorem.
We need to find \[ P(B|D) \] By Bayes' theorem, \[ P(B|D)=\frac{P(B)P(D|B)} {P(A)P(D|A)+P(B)P(D|B)+P(C)P(D|C)} \]
Step 3: Substitute the values.
\[ P(B|D) = \frac{\frac{2}{5}\cdot\frac{3}{100}} {\frac{3}{10}\cdot\frac{2}{100} +\frac{2}{5}\cdot\frac{3}{100} +\frac{3}{10}\cdot\frac{1}{100}} \] Now, \[ \frac{2}{5}\cdot\frac{3}{100} = \frac{6}{500} = \frac{3}{250} \] Also, \[ \frac{3}{10}\cdot\frac{2}{100} = \frac{6}{1000} \] \[ \frac{2}{5}\cdot\frac{3}{100} = \frac{12}{1000} \] \[ \frac{3}{10}\cdot\frac{1}{100} = \frac{3}{1000} \] Therefore, the denominator is \[ \frac{6}{1000}+\frac{12}{1000}+\frac{3}{1000} = \frac{21}{1000} \]
Step 4: Simplify the probability.
\[ P(B|D) = \frac{\frac{12}{1000}}{\frac{21}{1000}} \] \[ =\frac{12}{21} \] \[ =\frac{4}{7} \]
Step 5: Final conclusion.
Therefore, the probability that the defective toy was manufactured by machine \(B\) is \[ \boxed{\frac{4}{7}} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Probability Questions
S is the sample space and A, B are two events of a random experiment. Match the items of List A with the items of List B.
Then the correct match is:- If \( P(A \cap B) + P(B \mid A \cap B) = \), then:
- Two digits are selected at random from the digits 1 through 9. If their sum is even, then the probability that both are odd is:
For the probability distribution of a discrete random variable \( X \) as given below, the mean of \( X \) is:
- In a random experiment, two dice are thrown and the sum of the numbers appeared on them is recorded. This experiment is repeated 9 times. If the probability that a sum of 6 appears at least once is \( P_1 \) and a sum of 8 appears at least once is \( P_2 \), then \( P_1 : P_2 = \):
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions