Samacheer Kalvi Class 9 Maths Solutions Chapter 9 Probability Exercise 9.2

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.2

 

Question 1. A company manufactures 10000 Laptops in 6 months. In that 25 of them are found to be defective. When you choose one Laptop from the manufactured, what is the probability that selected Laptop is a good one?
Answer:
Total number of laptops \( n(S) = 10000 \)
Number of defective laptops = 25
Number of good laptops \( n(E) = 10000 - 25 = 9975 \)
The probability of selecting a good laptop is found by dividing the number of good laptops by the total number of laptops.
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{9975}{10000} \)
\( P(E) = \frac{399}{400} \)
\( P(E) = 0.9975 \)
In simple words: First, find how many laptops are good. Then, divide this number by the total number of laptops made. This gives you the chance of picking a good one.

🎯 Exam Tip: Remember that probability is always a value between 0 and 1 (inclusive). Make sure your final answer falls within this range.

 

Question 2. In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their voter ID card. If a youngster is selected at random, find the probability that the youngster does not have their voter ID card.
Answer:
Total number of youngsters surveyed \( n(S) = 400 \)
Number of youngsters having voter ID cards = 191
Number of youngsters who do not have voter ID cards \( n(E) = 400 - 191 = 209 \)
The probability of selecting a youngster without a voter ID card is calculated by dividing the number of those without IDs by the total number of youngsters.
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{209}{400} \)
Therefore, the required probability is \( \frac{209}{400} \).
In simple words: Find how many youngsters do not have a voter ID. Then, divide that number by the total number of youngsters surveyed to get the probability.

🎯 Exam Tip: For "not" events, calculate the number of outcomes for the opposite event and subtract from the total, or use the complement rule \( P(E') = 1 - P(E) \).

 

Question 3. The probability of guessing the correct answer to a certain question is \( \frac{x}{3} \). If the probability of not guessing the correct answer is \( \frac{x}{5} \), then find the value of x.
Answer:
Let E be the event of guessing the correct answer.
Probability of guessing the correct answer \( P(E) = \frac{x}{3} \)
Let E' be the event of not guessing the correct answer.
Probability of not guessing the correct answer \( P(E') = \frac{x}{5} \)
We know that the sum of the probability of an event and the probability of its complement is always 1.
\( P(E) + P(E') = 1 \)
\( \frac{x}{3} + \frac{x}{5} = 1 \)
To add the fractions, find a common denominator, which is 15.
\( \frac{5x}{15} + \frac{3x}{15} = 1 \)
\( \frac{5x+3x}{15} = 1 \)
\( \implies \frac{8x}{15} = 1 \)
To find x, multiply both sides by 15 and then divide by 8.
\( 8x = 15 \)
\( x = \frac{15}{8} \)
Thus, the value of x is \( \frac{15}{8} \).
In simple words: The chance of getting something right plus the chance of getting it wrong always adds up to 1. Use this rule with the given fractions to solve for 'x'.

🎯 Exam Tip: Remember the fundamental probability rule: \( P(A) + P(A') = 1 \). This allows you to find one probability if the other is known.

 

Question 4. If a probability of a player winning a particular tennis match is 0.72. What is the probability of the player loosing the match?
Answer:
Let E be the event that the player wins the tennis match.
Probability of a player winning a tennis match \( P(E) = 0.72 \)
Let E' be the event that the player loses the tennis match.
We know that the probability of an event happening plus the probability of it not happening is 1.
\( P(E) + P(E') = 1 \)
\( 0.72 + P(E') = 1 \)
To find the probability of losing, subtract the probability of winning from 1.
\( P(E') = 1 - 0.72 \)
\( P(E') = 0.28 \)
Therefore, the probability of the player losing the match is 0.28.
In simple words: If you know the chance of winning, subtract that chance from 1 to find the chance of losing.

🎯 Exam Tip: Ensure that your probabilities are always positive. A probability of 0 means impossible, and 1 means certain, with all other values in between.

 

Question 5. A survey recorded data about maids at homes as shown in the table below. A family is selected at random. Find the probability that the family selected has:

(i) Both types of maids
(ii) Part time maids
(iii) No maids
Answer:
Total number of families surveyed \( n(S) = 860 + 370 + 250 = 1480 \)
However, the problem states total families surveyed = 1500 (from the solution steps, not explicitly in the table data sum). We will use \( n(S) = 1500 \) as given in the solution.
Number of families used maids = 860 + 370 + 250 = 1480
Number of families not using any maids = \( 1500 - 1480 = 20 \)

(i) Let \( E_1 \) be the event of getting families that use both types of maids.
From the table, the number of families using both types of maids \( n(E_1) = 250 \).
The probability of a family having both types of maids is the number of such families divided by the total surveyed.
\( P(E_1) = \frac{n(E_1)}{n(S)} \)
\( P(E_1) = \frac{250}{1500} \)
\( P(E_1) = \frac{25}{150} \)
\( P(E_1) = \frac{1}{6} \)
The probability of getting both types of maids is \( \frac{1}{6} \).

(ii) Let \( E_2 \) be the event of getting families that use part-time maids.
From the table, the number of families using only part-time maids \( n(E_2) = 860 \).
The probability of a family having part-time maids is the number of such families divided by the total surveyed.
\( P(E_2) = \frac{n(E_2)}{n(S)} \)
\( P(E_2) = \frac{860}{1500} \)
\( P(E_2) = \frac{43}{75} \)
The probability of getting part-time maids is \( \frac{43}{75} \).

(iii) Let \( E_3 \) be the event of getting families that use no maids.
We calculated the number of families not using any maids as \( n(E_3) = 20 \).
The probability of a family having no maids is this number divided by the total surveyed.
\( P(E_3) = \frac{n(E_3)}{n(S)} \)
\( P(E_3) = \frac{20}{1500} \)
\( P(E_3) = \frac{2}{150} \)
\( P(E_3) = \frac{1}{75} \)
The probability of getting no maids is \( \frac{1}{75} \).
In simple words: For each type of maid situation, divide the number of families that fit that situation by the total number of families surveyed. Always simplify the fractions to their smallest form.

🎯 Exam Tip: When dealing with survey data, carefully identify the 'total outcomes' (\( n(S) \)) and the 'favorable outcomes' (\( n(E) \)) for each specific question before calculating probability.