Samacheer Kalvi Class 11 Maths Solutions Chapter 12 Introduction to Probability Theory Exercise 12.4

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 12 Introduction to Probability Theory Ex 12.4

 

Question 1. A factory has two Machines - I and II. Machines - I produce 60% of items and Machine - II produces 40 % of the items of the total output. Further 2 % of the items produced by Machine - I are defective whereas 4 % produced by Machine -II are defective. If an item is drawn at random what is the probability that it is defective?
Answer: Let \( A_1 \) be the event that items are produced by Machine – I, and \( A_2 \) be the event that items are produced by Machine – II.
Let B be the event of drawing a defective item.
We are given the following probabilities:
\( P(A_1) = 60\% = \frac{60}{100} = 0.60 \)
\( P(A_2) = 40\% = \frac{40}{100} = 0.40 \)

The conditional probabilities of a defective item are:
\( P(B/A_1) = 2\% = \frac{2}{100} = 0.02 \)
\( P(B/A_2) = 4\% = \frac{4}{100} = 0.04 \)

To find the total probability of event B (drawing a defective item), we use the law of total probability, since \( A_1 \) and \( A_2 \) are mutually exclusive and exhaustive events:
\( P(B) = P(A_1) \cdot P(B/A_1) + P(A_2) \cdot P(B/A_2) \)
\( P(B) = 0.60 \times 0.02 + 0.40 \times 0.04 \)
\( P(B) = 0.012 + 0.016 \)
\( P(B) = 0.028 \)
In simple words: We calculated the overall chance that a randomly picked item is defective. This considers how many items each machine makes and how often each machine produces defective items.

🎯 Exam Tip: Remember that the sum of probabilities for all possible outcomes (like \( P(A_1) + P(A_2) \)) must equal 1 in probability problems.

 

Question 2. There are two identical urns containing respectively 6 black and 4 red balls, 2 black, and 2 red balls. An urn is chosen at random and a ball is drawn from it.
(i) Find the probability that the ball is black
(ii) if the ball is black, what is the probability that it is from the first urn?
Answer: First, let's list the number of balls in each urn:

(i) Find the probability that the ball is black.
Let \( A_1 \) be the event of selecting Urn I, and \( A_2 \) be the event of selecting Urn II.
Since an urn is chosen at random, the probability of selecting each urn is:
\( P(A_1) = \frac{1}{2} \)
\( P(A_2) = \frac{1}{2} \)

Let B be the event of drawing a black ball.
Conditional probability of drawing a black ball from Urn I:
\( P(B/A_1) = \frac{6C_1}{10C_1} = \frac{6}{10} = \frac{3}{5} \)
Conditional probability of drawing a black ball from Urn II:
\( P(B/A_2) = \frac{2C_1}{4C_1} = \frac{2}{4} = \frac{1}{2} \)

Using the law of total probability to find \( P(B) \):
\( P(B) = P(A_1) \cdot P(B/A_1) + P(A_2) \cdot P(B/A_2) \)
\( P(B) = \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{1}{2} \)
\( P(B) = \frac{3}{10} + \frac{1}{4} \)
\( P(B) = \frac{6}{20} + \frac{5}{20} \)
\( P(B) = \frac{11}{20} \)

(ii) If the ball is black, what is the probability that it is from the first urn?
We need to find the conditional probability \( P(A_1/B) \) using Bayes' theorem:
\( P(A_1/B) = \frac{P(A_1) \cdot P(B/A_1)}{P(B)} \)
\( P(A_1/B) = \frac{\frac{1}{2} \times \frac{3}{5}}{\frac{11}{20}} \)
\( P(A_1/B) = \frac{\frac{3}{10}}{\frac{11}{20}} \)
\( P(A_1/B) = \frac{3}{10} \times \frac{20}{11} \)
\( P(A_1/B) = \frac{6}{11} \)
In simple words: First, we found the overall probability of drawing a black ball from either urn. Then, using Bayes' theorem, we calculated the chance that a black ball we already picked came specifically from the first urn. This helps us understand the source of an event.

🎯 Exam Tip: When using Bayes' theorem, clearly define all events and state the formula before plugging in the values. Pay attention to simplifying fractions correctly.

 

Question 3. A firm manufactures PVC pipes in the three plants viz, X, Y, and Z. The daily production volumes from the three firms X, Y, and Z are respectively 2000 units, 3000 units, and 5000 units. It is known from past experience that 3 % of the output from plant X, 4 % from plant Y, and 2 % from plant Z are defective. A pipe is selected at random from a day's total production
(i) find the probability that the selected pipe is a defective one?
(ii) if the selected pipe is defective, then what is the probability that it was produced by plant Y?
Answer: Let \( A_1 \) be the event that the pipe is produced by Plant X, \( A_2 \) by Plant Y, and \( A_3 \) by Plant Z.
Let B be the event that the selected pipe is defective.

The total production is \( 2000 + 3000 + 5000 = 10000 \) units.

Probabilities of a pipe coming from each plant:
\( P(A_1) = \frac{2000}{10000} = \frac{1}{5} \)
\( P(A_2) = \frac{3000}{10000} = \frac{3}{10} \)
\( P(A_3) = \frac{5000}{10000} = \frac{1}{2} \)

Conditional probabilities of a pipe being defective from each plant:
\( P(B/A_1) = 3\% = 0.03 \)
\( P(B/A_2) = 4\% = 0.04 \)
\( P(B/A_3) = 2\% = 0.02 \)

(i) Find the probability that the selected pipe is a defective one.
Using the law of total probability:
\( P(B) = P(A_1) \cdot P(B/A_1) + P(A_2) \cdot P(B/A_2) + P(A_3) \cdot P(B/A_3) \)
\( P(B) = \frac{1}{5} \times 0.03 + \frac{3}{10} \times 0.04 + \frac{1}{2} \times 0.02 \)
\( P(B) = 0.2 \times 0.03 + 0.3 \times 0.04 + 0.5 \times 0.02 \)
\( P(B) = 0.006 + 0.012 + 0.010 \)
\( P(B) = 0.028 \)
This can also be expressed as \( P(B) = \frac{28}{1000} = \frac{7}{250} \).

(ii) If the selected pipe is defective, then what is the probability that it was produced by plant Y?
We need to find \( P(A_2/B) \) using Bayes' theorem:
\( P(A_2/B) = \frac{P(A_2) \cdot P(B/A_2)}{P(B)} \)
\( P(A_2/B) = \frac{\frac{3}{10} \times 0.04}{\frac{7}{250}} \)
\( P(A_2/B) = \frac{0.3 \times 0.04}{0.028} \)
\( P(A_2/B) = \frac{0.012}{0.028} \)
To simplify, multiply numerator and denominator by 1000:
\( P(A_2/B) = \frac{12}{28} \)
Dividing both by 4:
\( P(A_2/B) = \frac{3}{7} \)
In simple words: First, we calculated the overall chance of picking a bad pipe from all factories. Then, if we find a defective pipe, we used Bayes' theorem to figure out how likely it is that this specific bad pipe came from Plant Y. This helps factories identify problem areas more effectively.

🎯 Exam Tip: Always make sure your total probability (denominator in Bayes' theorem) includes contributions from all relevant events. This ensures a complete and accurate calculation.

 

Question 4. The chances of A, B, and C becoming manager of a certain company are 5 : 3 : 2. The probabilities that the office canteen will be improved if A, B, and C become managers are 0.4, 0.5 and 0.3 respectively. If the office canteen has been improved, what is the probability that B was appointed as the manager?
Answer: Let \( A_1 \) be the event that A becomes manager, \( A_2 \) that B becomes manager, and \( A_3 \) that C becomes manager.
Let B be the event that the office canteen is improved.

The chances of becoming manager are in the ratio 5:3:2. The total sum of ratio parts is \( 5+3+2 = 10 \).
So, the probabilities are:
\( P(A_1) = \frac{5}{10} \)
\( P(A_2) = \frac{3}{10} \)
\( P(A_3) = \frac{2}{10} \)

The probabilities of canteen improvement given each manager are:
\( P(B/A_1) = 0.4 \)
\( P(B/A_2) = 0.5 \)
\( P(B/A_3) = 0.3 \)

We need to find \( P(A_2/B) \), the probability that B was appointed as manager given that the canteen has been improved. We will use Bayes' theorem.

First, calculate the total probability of the canteen being improved, \( P(B) \), using the law of total probability:
\( P(B) = P(A_1) \cdot P(B/A_1) + P(A_2) \cdot P(B/A_2) + P(A_3) \cdot P(B/A_3) \)
\( P(B) = \frac{5}{10} \times 0.4 + \frac{3}{10} \times 0.5 + \frac{2}{10} \times 0.3 \)
\( P(B) = 0.5 \times 0.4 + 0.3 \times 0.5 + 0.2 \times 0.3 \)
\( P(B) = 0.20 + 0.15 + 0.06 \)
\( P(B) = 0.41 \)

Now, apply Bayes' theorem to find \( P(A_2/B) \):
\( P(A_2/B) = \frac{P(A_2) \cdot P(B/A_2)}{P(B)} \)
\( P(A_2/B) = \frac{\frac{3}{10} \times 0.5}{0.41} \)
\( P(A_2/B) = \frac{0.3 \times 0.5}{0.41} \)
\( P(A_2/B) = \frac{0.15}{0.41} \)
To remove decimals, we multiply the numerator and denominator by 100:
\( P(A_2/B) = \frac{15}{41} \)
In simple words: We calculated the chance that B became the manager, knowing that the office canteen improvement has already happened. This involves first finding the overall chance of canteen improvement and then using that to figure out the specific manager's probability.

🎯 Exam Tip: When given probabilities as ratios, always convert them to fractions (out of the total sum of the ratio) before using them in calculations to avoid errors.

 

Question 5. An advertising executive is studying television viewing habits of married men and women during prime time hours. Based on the past viewing records he has determined that during prime time wives are watching television 60 % of the time. It has also been determined that when the wife is watching television, 40 % of the time the husband is also watching. When the wife is not watching the television, 30% of the time the husband is watching the television. Find the probability that
(i) the husband is watching the television during the prime time of television
(ii) if the husband is watching the television, the wife is also watching the television.
Answer: Let W be the event that the wife is watching television.
Let \( W' \) be the event that the wife is not watching television.
Let H be the event that the husband is watching television.

Given probabilities:
\( P(W) = 60\% = \frac{60}{100} \)
So, the probability that the wife is not watching is \( P(W') = 1 - P(W) = 1 - \frac{60}{100} = \frac{40}{100} \)

Conditional probabilities of the husband watching:
\( P(H/W) = 40\% = \frac{40}{100} \) (probability husband watches given wife is watching)
\( P(H/W') = 30\% = \frac{30}{100} \) (probability husband watches given wife is not watching)

(i) Find the probability that the husband is watching the television during prime time.
We need to find \( P(H) \). Using the law of total probability:
\( P(H) = P(H/W) \cdot P(W) + P(H/W') \cdot P(W') \)
\( P(H) = \frac{40}{100} \times \frac{60}{100} + \frac{30}{100} \times \frac{40}{100} \)
\( P(H) = \frac{2400}{10000} + \frac{1200}{10000} \)
\( P(H) = \frac{24}{100} + \frac{12}{100} \)
\( P(H) = \frac{36}{100} = \frac{9}{25} \)

(ii) If the husband is watching the television, find the probability that the wife is also watching the television.
We need to find \( P(W/H) \). Using Bayes' theorem:
\( P(W/H) = \frac{P(H/W) \cdot P(W)}{P(H)} \)
\( P(W/H) = \frac{\frac{40}{100} \times \frac{60}{100}}{\frac{36}{100}} \)
\( P(W/H) = \frac{\frac{24}{100}}{\frac{36}{100}} \)
\( P(W/H) = \frac{24}{36} \)
Simplifying the fraction by dividing both by 12:
\( P(W/H) = \frac{2}{3} \)
In simple words: First, we calculated the total chance that a husband would be watching TV during prime time, considering whether his wife was also watching or not. Then, if we know the husband is watching TV, we found out the probability that his wife is watching too. This helps understand viewing patterns better.

🎯 Exam Tip: Always clearly define your events (like W, W', H) and their probabilities at the beginning of the solution. This prevents confusion, especially in conditional probability problems.