Samacheer Kalvi Class 9 Maths Solutions Chapter 9 Probability Exercise 9.1

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 9 Probability Ex 9.1

Question 1. You are walking along a street. If you just choose a stranger crossing you, what is the probability that his next birthday will fall on a Sunday?
Answer: The sample space (S) for days of the week is {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}.
So, the total number of possible outcomes, n(S), is 7.
Let E be the event that the birthday falls on a Sunday. There is only one Sunday in a week, so n(E) = 1.
The probability P(E) is calculated as the number of favorable outcomes divided by the total number of outcomes.
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{1}{7} \)
In simple words: There are 7 days in a week. Only one of them is Sunday. So, the chance of a random birthday being on a Sunday is 1 out of 7.

🎯 Exam Tip: When dealing with daily events, remember there are 7 days in a week. For simple probabilities like this, identify the total possible outcomes and the specific outcomes you are looking for.

Question 2. What is the probability of drawing a King or a Queen or a Jack from a deck of cards?
Answer: In a standard deck of 52 playing cards, the total number of possible outcomes, n(S), is 52.
Let E be the event of drawing a King or a Queen or a Jack.
There are 4 Kings, 4 Queens, and 4 Jacks in a deck of cards. Each suit has one of each.
The number of favorable outcomes, n(E), is the sum of these cards: \( n(E) = 4 + 4 + 4 = 12 \).
The probability P(E) is calculated as:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{12}{52} \)
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4.
\( P(E) = \frac{12 \div 4}{52 \div 4} = \frac{3}{13} \)
In simple words: There are 52 cards in total. If you want a King, Queen, or Jack, there are 12 such cards. So, the chance of picking one of these is 12 out of 52, which simplifies to 3 out of 13.

🎯 Exam Tip: Always simplify fractions to their lowest terms in probability answers. Remember the composition of a standard 52-card deck (4 suits, 13 cards per suit).

Question 3. What is the probability of throwing an even number with a single standard dice of six faces?
Answer: When a single standard six-sided die is rolled, the sample space (S) consists of the numbers {1, 2, 3, 4, 5, 6}.
The total number of possible outcomes, n(S), is 6.
Let E be the event of throwing an even number.
The even numbers in the sample space are {2, 4, 6}.
The number of favorable outcomes, n(E), is 3.
The probability P(E) is calculated as:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{3}{6} \)
We can simplify this fraction by dividing both the numerator and the denominator by 3.
\( P(E) = \frac{3 \div 3}{6 \div 3} = \frac{1}{2} \)
In simple words: A regular dice has 6 sides, and 3 of those sides have even numbers (2, 4, 6). So, the chance of rolling an even number is 3 out of 6, which is the same as 1 out of 2.

🎯 Exam Tip: Understand that "standard dice" usually means a six-sided cube with faces numbered 1 to 6. Clearly list the favorable outcomes for each event before calculating probability.

Question 4. There are 24 balls in a pot. If 3 of them are Red, 5 of them are Blue and the remaining are Green then, what is the probability of picking out (i) a Blue ball, (ii) a Red ball and (iii) a Green ball?
Answer: The total number of balls in the pot is 24. So, the total number of possible outcomes, n(S), is 24.
Number of Red balls = 3
Number of Blue balls = 5
The remaining balls are Green. To find the number of Green balls, subtract the number of Red and Blue balls from the total.
Number of Green balls = \( 24 - (3 + 5) \)
Number of Green balls = \( 24 - 8 \)
Number of Green balls = 16

(i) Let \( E_1 \) be the event of picking a Blue ball.
The number of Blue balls, \( n(E_1) \), is 5.
The probability \( P(E_1) \) of picking a Blue ball is:
\( P(E_1) = \frac{n(E_1)}{n(S)} \)
\( P(E_1) = \frac{5}{24} \)

(ii) Let \( E_2 \) be the event of picking a Red ball.
The number of Red balls, \( n(E_2) \), is 3.
The probability \( P(E_2) \) of picking a Red ball is:
\( P(E_2) = \frac{n(E_2)}{n(S)} \)
\( P(E_2) = \frac{3}{24} \)
This fraction can be simplified by dividing both parts by 3.
\( P(E_2) = \frac{3 \div 3}{24 \div 3} = \frac{1}{8} \)

(iii) Let \( E_3 \) be the event of picking a Green ball.
The number of Green balls, \( n(E_3) \), is 16.
The probability \( P(E_3) \) of picking a Green ball is:
\( P(E_3) = \frac{n(E_3)}{n(S)} \)
\( P(E_3) = \frac{16}{24} \)
This fraction can be simplified by dividing both parts by 8.
\( P(E_3) = \frac{16 \div 8}{24 \div 8} = \frac{2}{3} \)
In simple words: First, find out how many green balls there are. Then, for each color (blue, red, green), divide the number of balls of that color by the total number of balls (24). This gives you the chance of picking that color. Always simplify the fractions if you can.

🎯 Exam Tip: Always make sure to calculate the total number of items correctly. When dealing with "remaining" items, subtract the known counts from the total. Remember to simplify all probability fractions.

Question 5. When two coins are tossed, what is the probability that two heads are obtained?
Answer: When two coins are tossed, the possible outcomes are Head-Head (HH), Head-Tail (HT), Tail-Head (TH), and Tail-Tail (TT).
So, the sample space (S) is {(H, H), (H, T), (T, H), (T, T)}.
The total number of possible outcomes, n(S), is 4.
Let E be the event of getting two heads.
The only outcome with two heads is (H, H).
The number of favorable outcomes, n(E), is 1.
The probability P(E) is calculated as:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{1}{4} \)
In simple words: If you flip two coins, there are four possible results. Only one of these results is getting two heads. So, the chance is 1 out of 4.

🎯 Exam Tip: For coin toss problems, carefully list all possible outcomes to determine the sample space. This helps in correctly counting both total outcomes and favorable outcomes.

Question 6. Two dice are rolled, find the probability that the sum is
(i) equal to 1
(ii) equal to 4
(iii) less than 13
Answer: When two dice are rolled, each die has 6 faces. The total number of possible outcomes is \( 6 \times 6 = 36 \).
The sample space (S) consists of all pairs from (1,1) to (6,6). For example:
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
So, the total number of possible outcomes, n(S), is 36.

(i) Let \( E_1 \) be the event that the sum of the numbers is equal to 1.
The smallest possible sum when rolling two dice is \( 1+1=2 \). It is not possible to get a sum of 1.
Therefore, the number of favorable outcomes, \( n(E_1) \), is 0.
The probability \( P(E_1) \) is:
\( P(E_1) = \frac{n(E_1)}{n(S)} \)
\( P(E_1) = \frac{0}{36} = 0 \)

(ii) Let \( E_2 \) be the event that the sum of the numbers is equal to 4.
The pairs that sum to 4 are {(1, 3), (2, 2), (3, 1)}.
The number of favorable outcomes, \( n(E_2) \), is 3.
The probability \( P(E_2) \) is:
\( P(E_2) = \frac{n(E_2)}{n(S)} \)
\( P(E_2) = \frac{3}{36} \)
This fraction can be simplified by dividing both parts by 3.
\( P(E_2) = \frac{3 \div 3}{36 \div 3} = \frac{1}{12} \)

(iii) Let \( E_3 \) be the event that the sum of the numbers is less than 13.
The maximum possible sum when rolling two dice is \( 6+6=12 \).
Since all possible sums (from 2 to 12) are less than 13, every outcome in the sample space is a favorable outcome.
Therefore, the number of favorable outcomes, \( n(E_3) \), is 36.
The probability \( P(E_3) \) is:
\( P(E_3) = \frac{n(E_3)}{n(S)} \)
\( P(E_3) = \frac{36}{36} = 1 \)
In simple words: When you roll two dice, there are 36 possible results. For a sum of 1, it's impossible, so the chance is 0. For a sum of 4, there are 3 ways to get it, so the chance is 3 out of 36 (or 1 in 12). For a sum less than 13, all results are less than 13, so the chance is 1 (or 100%).

🎯 Exam Tip: Always list out the sample space clearly, especially for two-dice problems, to avoid missing outcomes. Remember that a probability of 0 means an impossible event, and a probability of 1 means a certain event.

Question 7. A manufacturer tested 7000 LED lights at random and found that 25 of them were defective. If a LED light is selected at random, what is the probability that the selected LED light is a defective one.
Answer: The manufacturer tested 7000 LED lights. This is the total number of possible outcomes, n(S) = 7000.
It was found that 25 of these LED lights were defective.
Let \( E_1 \) be the event that a selected LED light is defective.
The number of favorable outcomes (defective lights), \( n(E_1) \), is 25.
The probability \( P(E_1) \) of selecting a defective LED light is:
\( P(E_1) = \frac{n(E_1)}{n(S)} \)
\( P(E_1) = \frac{25}{7000} \)
To simplify this fraction, divide both the numerator and denominator by 25.
\( P(E_1) = \frac{25 \div 25}{7000 \div 25} = \frac{1}{280} \)
In simple words: Out of 7000 lights, 25 were broken. So, the chance of picking a broken light is 25 out of 7000, which simplifies to 1 out of 280.

🎯 Exam Tip: For quality control problems, clearly identify the total number of items tested and the number of items exhibiting the specific characteristic (e.g., defective) to calculate the probability accurately.

Question 8. In a football match, a goalkeeper of a team can stop the goal, 32 times out of 40 attempts tried by a team. Find the probability that the opponent team can convert the attempt into a goal.
Answer: The total number of attempts made by the opponent team is 40. This is the total number of possible outcomes, n(S) = 40.
The goalkeeper stopped 32 of these attempts.
We need to find the probability that the opponent team *converts* an attempt into a goal. This means the attempts that were *not* stopped by the goalkeeper.
Number of attempts converted into a goal = Total attempts - Attempts stopped by goalkeeper
Number of attempts converted into a goal = \( 40 - 32 = 8 \)
Let E be the event that the opponent team converts an attempt into a goal.
The number of favorable outcomes, n(E), is 8.
The probability P(E) is calculated as:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{8}{40} \)
This fraction can be simplified by dividing both the numerator and the denominator by 8.
\( P(E) = \frac{8 \div 8}{40 \div 8} = \frac{1}{5} \)
In simple words: There were 40 attempts, and 32 were blocked. This means 8 attempts became goals. So, the chance of the opponent scoring a goal is 8 out of 40, which simplifies to 1 out of 5.

🎯 Exam Tip: Read the question carefully to determine if you need to calculate the probability of an event happening or the probability of it *not* happening (its complement). Here, "convert into a goal" is the complement of "stop the goal."

Question 9. What is the probability that the spinner will not land on a multiple of 3?

Answer: The spinner has 8 equal parts, numbered from 1 to 8. So, the sample space (S) is {1, 2, 3, 4, 5, 6, 7, 8}.
The total number of possible outcomes, n(S), is 8.
First, identify the numbers that are multiples of 3 within the sample space. These are {3, 6}.
The number of outcomes that are multiples of 3 is 2.
We need to find the probability that the spinner will *not* land on a multiple of 3.
Number of outcomes that are not multiples of 3 = Total outcomes - Number of multiples of 3
Number of outcomes that are not multiples of 3 = \( 8 - 2 = 6 \).
Let E be the event that the spinner does not land on a multiple of 3.
The number of favorable outcomes, n(E), is 6.
The probability P(E) is calculated as:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{6}{8} \)
This fraction can be simplified by dividing both parts by 2.
\( P(E) = \frac{6 \div 2}{8 \div 2} = \frac{3}{4} \)
In simple words: The spinner has numbers 1 to 8. The numbers 3 and 6 are multiples of 3. There are 6 numbers that are not multiples of 3 (1, 2, 4, 5, 7, 8). So, the chance of not landing on a multiple of 3 is 6 out of 8, which is 3 out of 4.

🎯 Exam Tip: When asked for the probability of "not" an event, it's often easier to calculate the probability of the event *itself* happening and subtract it from 1 (or subtract the number of favorable outcomes from the total). Ensure all fractions are simplified.

Question 10. Frame two problems in calculating probability, based on the spinner shown here.

Answer: Here are two probability problems based on the given spinner:
(i) What is the probability that the spinner will land on an odd number?
(ii) What is the probability that the spinner will not land on a multiple of 2?
In simple words: Based on the spinner that has numbers from 1 to 8, we can ask questions like "What are the chances of landing on an odd number?" or "What are the chances of not landing on a number that can be divided by 2?"

🎯 Exam Tip: When framing probability problems, ensure they are clearly worded and use terms like "probability," "chance," or "likelihood." The problems should be solvable using the provided data (the spinner numbers). Choose events that are distinct and cover different probability concepts.