Samacheer Kalvi Class 9 Maths Solutions Chapter 9 Probability More Ques

I. Choose the Correct Answer

Question 1. Which of the following cannot be taken as probability of an event?
(a) 0
(b) 0.35
(c) \( \frac{7}{20} \)
(d) \( -\frac{7}{20} \)
Answer: (d) \( -\frac{7}{20} \)
In simple words: Probability values must always be between 0 and 1, inclusive. A negative value for probability is not possible, as it represents a chance that cannot exist.

🎯 Exam Tip: Remember that probability can never be less than 0 (impossible event) or greater than 1 (certain event); it always falls within the range [0, 1].

Question 2. A letter is chosen at random from the word "MATHEMATICS” the probability of getting a vowel is ..........
(a) \( \frac{2}{11} \)
(b) \( \frac{3}{11} \)
(c) \( \frac{4}{11} \)
(d) \( \frac{5}{11} \)
Answer: (c) \( \frac{4}{11} \)
In simple words: The word "MATHEMATICS" has 11 letters in total. The vowels are A, E, A, I, so there are 4 vowels. The chance of picking a vowel is the number of vowels divided by the total number of letters.

🎯 Exam Tip: When counting letters for probability, remember to count repeated letters as separate outcomes, and be careful to identify all vowels (A, E, I, O, U).

Question 3. If \( P(A) = \frac{1}{3} \) then \( P(A') \) is ..........
(a) \( \frac{1}{3} \)
(b) \( \frac{2}{3} \)
(c) \( \frac{1}{2} \)
(d) 1
Answer: (b) \( \frac{2}{3} \)
In simple words: The probability of an event happening plus the probability of it not happening always adds up to 1. If an event has a \( \frac{1}{3} \) chance of happening, it has a \( \frac{2}{3} \) chance of not happening.

🎯 Exam Tip: Always use the formula \( P(A') = 1 - P(A) \) to find the probability of a complementary event, ensuring your calculations are accurate.

Question 4. An integer is chosen from the first twenty natural number, the probability that it is a prime number is ..........
(a) \( \frac{1}{5} \)
(b) \( \frac{2}{5} \)
(c) \( \frac{3}{5} \)
(d) \( \frac{4}{5} \)
Answer: (b) \( \frac{2}{5} \)
In simple words: From the numbers 1 to 20, we need to find how many are prime numbers. Prime numbers are those greater than 1 that only have two factors: 1 and themselves. The prime numbers are 2, 3, 5, 7, 11, 13, 17, 19. There are 8 prime numbers out of 20 total numbers, which simplifies to \( \frac{2}{5} \).

🎯 Exam Tip: Carefully list all prime numbers within the given range and count them accurately; remember that 1 is not a prime number.

Question 5. From a well shuffled pack of 52 cards one card is drawn at random. The probability of getting not a king is ..........
(a) \( \frac{12}{13} \)
(b) \( \frac{1}{13} \)
(c) \( \frac{4}{13} \)
(d) \( \frac{2}{13} \)
Answer: (a) \( \frac{12}{13} \)
In simple words: A standard deck of 52 cards has 4 kings. The chance of drawing a king is \( \frac{4}{52} \), which simplifies to \( \frac{1}{13} \). To find the probability of *not* getting a king, subtract this from 1. So, \( 1 - \frac{1}{13} = \frac{12}{13} \).

🎯 Exam Tip: Always remember that a standard deck has 52 cards and 4 suits; when calculating "not" probabilities, it's often easier to find the probability of the event itself and subtract it from 1.

II. Answer the Following Questions

Question 6. 1500 families with 2 children were selected randomly and the following data were recorded.

Compute the probability of a family chosen at random having one girl.
Answer:
Total number of families (Sample Space), \( n(S) = 1500 \).
Let E be the event that a family has one girl.
From the table, the number of families with one girl (Favorable Outcomes), \( n(E) = 814 \).
The probability of event E is calculated as:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{814}{1500} \)
We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2.
\( P(E) = \frac{407}{750} \)
In simple words: To find the chance of a family having one girl, we divide the number of families with one girl (814) by the total number of families surveyed (1500). This gives us a fraction that can be simplified.

🎯 Exam Tip: Always state the total number of outcomes (sample space) and the number of favorable outcomes clearly before applying the probability formula. Remember to simplify fractions to their lowest terms.

Question 7. The record of weather station shows that out of the part 250 consecutive days its whether forecast were correct 175 time. What is the probability that it was not correct on a given data?
Answer:
Total number of consecutive days, \( n(S) = 250 \).
Let E be the event that the weather forecast was correct.
Number of times the forecast was correct, \( n(E) = 175 \).
The probability that the forecast was correct is:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{175}{250} \)
Simplify the fraction by dividing both numerator and denominator by 25:
\( P(E) = \frac{7}{10} \)
Now, we need to find the probability that the forecast was *not* correct. This is the complement of event E, denoted as \( P(E') \).
\( P(E') = 1 - P(E) \)
\( P(E') = 1 - \frac{7}{10} \)
To subtract, find a common denominator:
\( P(E') = \frac{10}{10} - \frac{7}{10} \)
\( P(E') = \frac{10-7}{10} \)
\( P(E') = \frac{3}{10} \)
In simple words: First, we find the chance that the weather forecast was right. Out of 250 days, it was right 175 times. So, the chance it was right is \( \frac{175}{250} \). Then, to find the chance it was wrong, we subtract the "right" chance from 1.

🎯 Exam Tip: For "not" probabilities, it's often simpler to calculate the probability of the event happening and then subtract it from 1, rather than counting all negative outcomes.

Question 8. If A coin is tossed 200 times and is found that a tail comes up for 120 times. Find the probability of getting a tail.
Answer:
Total number of trials (coin tosses), \( n(S) = 200 \).
Let E be the event of getting a tail.
Number of times a tail comes up (favorable outcomes), \( n(E) = 120 \).
The probability of getting a tail is:
\( P(E) = \frac{n(E)}{n(S)} \)
\( P(E) = \frac{120}{200} \)
Simplify the fraction by dividing both numerator and denominator by 20:
\( P(E) = \frac{6}{10} \)
Further simplify by dividing both by 2:
\( P(E) = \frac{3}{5} \)
In simple words: We tossed a coin 200 times. Tails showed up 120 times. So, the chance of getting a tail is the number of tails divided by the total number of tosses. This fraction can be made simpler.

🎯 Exam Tip: Ensure that when simplifying fractions for probability, you divide both the numerator and denominator by their greatest common divisor to get the simplest form.

Question 9. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball from the bag is thrice that of drawing a red ball, then find the number of blue balls in the bag.
Answer:
Let the number of blue balls be 'x'.
Number of red balls = 5.
Total number of balls in the bag, \( n(S) = 5 + x \).
Let B be the event of drawing a blue ball.
Let R be the event of drawing a red ball.
The probability of drawing a blue ball is \( P(B) = \frac{n(B)}{n(S)} = \frac{x}{5+x} \).
The probability of drawing a red ball is \( P(R) = \frac{n(R)}{n(S)} = \frac{5}{5+x} \).
According to the problem, the probability of drawing a blue ball is thrice that of drawing a red ball:
\( P(B) = 3P(R) \)

\( \implies \frac{x}{5+x} = 3 \times \frac{5}{5+x} \)

\( \implies \frac{x}{5+x} = \frac{15}{5+x} \)
Since the denominators are the same, the numerators must be equal.

\( \implies x = 15 \)
So, the number of blue balls in the bag is 15.
In simple words: We are told that the chance of picking a blue ball is three times more than picking a red ball. We set up an equation using 'x' for the number of blue balls. By solving this equation, we found that there are 15 blue balls in the bag.

🎯 Exam Tip: When dealing with probabilities involving unknown quantities, define a variable for the unknown, set up the probabilities as fractions, and then form an equation based on the given relationship to solve for the variable.

Question 10. Find the probability that a non leap year selected at random will have 53 fridays.
Answer:
A non-leap year has 365 days.
We know that 1 year = 52 weeks and 1 day.
So, a non-leap year has 52 full Fridays, and one extra day.
This one extra day can be any of the seven days of the week: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
The sample space (all possible outcomes for the extra day) is:
\( S = \{ \text{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} \} \)
So, the total number of outcomes, \( n(S) = 7 \).
Let A be the event that this extra day is a Friday.
Number of favorable outcomes for A, \( n(A) = 1 \) (since there is only one Friday in the list of possible extra days).
The probability that a non-leap year has 53 Fridays is:
\( P(A) = \frac{n(A)}{n(S)} \)
\( P(A) = \frac{1}{7} \)
In simple words: A regular year has 365 days, which means 52 full weeks and one extra day. For the year to have 53 Fridays, that extra day must be a Friday. Since the extra day can be any one of the seven days of the week, the chance of it being a Friday is 1 out of 7.

🎯 Exam Tip: Clearly define the sample space for the extra days in a non-leap year, and remember that only one of these days can be a Friday to result in 53 Fridays for the year.