Home
Class 12
MATHS
For A,B and C, the chances of being sele...

For A,B and C, the chances of being selected as the manager of a firm are `4:1:2` respectively. The probabilites for them to introduce a radical change in the marketing strategy are `0.3,0.8` and `0.5` respectively. If a change takes place, find the probability that it is due to the appoinment of B.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use Bayes' theorem. Let's break down the steps: ### Step 1: Define the Events Let: - \( E_1 \): Event that A is selected as the manager. - \( E_2 \): Event that B is selected as the manager. - \( E_3 \): Event that C is selected as the manager. - \( E \): Event that a radical change occurs in the marketing strategy. ### Step 2: Calculate the Probabilities of Selection The chances of being selected as the manager are given in the ratio \( 4:1:2 \). Thus, we can find the probabilities: - Total parts = \( 4 + 1 + 2 = 7 \) So, - \( P(E_1) = \frac{4}{7} \) - \( P(E_2) = \frac{1}{7} \) - \( P(E_3) = \frac{2}{7} \) ### Step 3: Probabilities of Introducing Change The probabilities for A, B, and C to introduce a radical change in the marketing strategy are given as: - \( P(E | E_1) = 0.3 \) - \( P(E | E_2) = 0.8 \) - \( P(E | E_3) = 0.5 \) ### Step 4: Apply Bayes' Theorem We need to find the probability that the change is due to the appointment of B, given that a change has occurred, i.e., \( P(E_2 | E) \). Using Bayes' theorem: \[ P(E_2 | E) = \frac{P(E | E_2) \cdot P(E_2)}{P(E)} \] ### Step 5: Calculate \( P(E) \) We need to calculate \( P(E) \) using the law of total probability: \[ P(E) = P(E | E_1) \cdot P(E_1) + P(E | E_2) \cdot P(E_2) + P(E | E_3) \cdot P(E_3) \] Substituting the values: \[ P(E) = (0.3 \cdot \frac{4}{7}) + (0.8 \cdot \frac{1}{7}) + (0.5 \cdot \frac{2}{7}) \] \[ = \frac{1.2}{7} + \frac{0.8}{7} + \frac{1.0}{7} = \frac{3.0}{7} \] ### Step 6: Substitute Values into Bayes' Theorem Now substitute back into Bayes' theorem: \[ P(E_2 | E) = \frac{P(E | E_2) \cdot P(E_2)}{P(E)} = \frac{0.8 \cdot \frac{1}{7}}{\frac{3.0}{7}} \] \[ = \frac{0.8}{3.0} = \frac{8}{30} = \frac{4}{15} \] ### Final Answer Thus, the probability that the change is due to the appointment of B is: \[ \boxed{\frac{4}{15}} \]
Promotional Banner

Topper's Solved these Questions

  • MATHEMATICS - 2013

    ICSE|Exercise Section-C|6 Videos
  • MATHEMATICS - 2013

    ICSE|Exercise Section-C|6 Videos
  • LINEAR REGRESSION

    ICSE|Exercise MULTIPLE CHOICE QUESTIONS|10 Videos
  • MATHEMATICS - 2014

    ICSE|Exercise Section-C|6 Videos

Similar Questions

Explore conceptually related problems

For A, B and C the chances of being selected as the manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities fro them to introduce a radical change in marketing strategy are 0.3, 0.8, and 0.5. if the change does take place, find the probability that it is due to the appointment of B or C.

Three persons A, B and C apply for a job of Manager in a Private Company. Chances of their selection (A, B and C) are in the ratio 1 : 2 : 4. The probabilities that A, B and C can introduce changes to improve profits of the company are 0.8, 0.5 and 0.3 respectively. If the change does not take place, find the probability that it is due to the appointment of C.

The probabilities of hitting a target by A, B and C are 3/5,3/4 and 1/3 respectively. If all three hits the target simultaneously then find the probability of hitting the target by the least two of them.

The probabilities of A, B and C solving a problem are 1/3, 2/7 and 3/8 respectively. If all the thee try to solve the problem simultaneously, find the probability that exactly one of them can solve it.

The probability of A, B and C solving a problem are (1)/(3) , (2)/(7) and (3)/(8) , respectively. If all try and solve the problem simultaneously, find the probability that only one of them will solve it.

A company has estimated that the probabilities of success fro there products introduced in the market are 1/3,2/5a n d2/3 respectively. Assuming independence, find the probability that the three products are successful. none of the products is successful,

Three persons A,B and C fire a target in turn . Their probabilities of hitting the target are 0.4 , 0.3 and 0.2 respectively . The probability of two hits is

The probabilities that the events A and B occur are 0.3 and 0.4 respectively. The probability that both A and B occur simultaneously is 0.15. What is the probability that neither A nor B occurs?

The probability that a student will receive A, B, C or D grade or 0.40, 0.3, 0.1 and 0.10 respectively. Find the probability that a student will receive. i. B or C grade ii. at most C grade.

The probability of two events A and B are 0.25 and 0.50 respectively. The probability of their simultaneous occurrences 0.15. Find the probability that neither A nor B occurs.