Samacheer Class 9 Maths Probability Book Back Answers & Solutions

Q: Two dice are rolled

(i) 0 (ii) \tfrac{1}{12} (iii) 1

Q: Frame two probability problems using the spinner

Example 1: P(even) = 1/2. Example 2: P(prime) = 1/2.

Your Progress - Chapter 90% complete

1Exercise 9.1 - Classical and Empirical Probability10 questions

Q.1Probability that a stranger’s next birthday falls on a Sundayv

Solution

There are 7 equally likely days; only Sunday is favourable. P = 1/7.

Answer:

\tfrac{1}{7}

Q.2Probability of drawing a King or Queen or Jack from a deck of cardsv

Solution

Favourable cards = 4 Kings + 4 Queens + 4 Jacks = 12. Total = 52. P = 12/52 = 3/13.

Answer:

\tfrac{3}{13}

Q.3Probability of throwing an even number on a diev

Solution

Even faces = {2,4,6} (3 outcomes). Total outcomes = 6. P = 3/6 = 1/2.

Answer:

\tfrac{1}{2}

Q.4Balls in a potv

Solution

Total balls = 24. (i) Blue = 5 → P = 5/24. (ii) Red = 3 → P = 3/24 = 1/8. (iii) Green = 24-3-5=16 → P = 16/24 = 2/3.

Answer:

(i) \tfrac{5}{24} (ii) \tfrac{1}{8} (iii) \tfrac{2}{3}

Q.5Probability of getting two heads when two coins are tossedv

Solution

Sample space = {HH, HT, TH, TT} (4 outcomes). Only HH is favourable. P = 1/4.

Answer:

\tfrac{1}{4}

Q.6Two dice are rolledv

Solution

Total outcomes = 36. (i) Sum = 1 is impossible → P = 0. (ii) Sum = 4: favourable = (1,3),(2,2),(3,1) → 3/36 = 1/12. (iii) Sum < 13: all sums 2..12 satisfy → 36/36 = 1.

Answer:

(i) 0 (ii) \tfrac{1}{12} (iii) 1

Q.7Probability of selecting defective LEDv

Solution

Total LEDs = 7000, defective = 25. P = 25/7000 = 1/280.

Answer:

\tfrac{1}{280}

Q.8Goalkeeper stops goal 32 times out of 40 attemptsv

Solution

Goals conceded = 40−32 = 8. Probability that a shot is a goal = 8/40 = 1/5.

Answer:

\tfrac{1}{5}

Q.9Probability that spinner will NOT land on a multiple of 3v

Solution

Spinner has 8 equal sectors; multiples of 3 = 2 sectors → non-multiples = 6. P = 6/8 = 3/4.

Answer:

\tfrac{3}{4}

Q.10Frame two probability problems using the spinnerv

Solution

Assume spinner sectors labeled 1–8 (equal). Problem 1: Find P(spinner lands on an even number). Even = {2,4,6,8} → 4/8 = 1/2. Problem 2: Find P(spinner lands on a prime number). Primes ≤8: {2,3,5,7} → 4/8 = 1/2.

Answer:

Example 1: P(even) = 1/2. Example 2: P(prime) = 1/2.

2Exercise 9.2 - Exercise 9.25 questions

Q.1Probability of selecting a good Laptopv

Solution

Total = 10000, defective = 25 → good = 10000−25 = 9975. P = 9975/10000 = 0.9975.

Answer:

0.9975

Q.2Probability that a youngster does NOT have voter IDv

Solution

Total surveyed = 400; having ID = 191 → without ID = 400−191 = 209. P = 209/400.

Answer:

\tfrac{209}{400}

Q.3Probability of guessing correctly is x/3 and probability of not guessing correctly is x/5; find xv

Solution

Assuming these two probabilities are complementary: x/3 + x/5 = 1. LCM 15: (5x+3x)/15 =1 → 8x/15 =1 → x = 15/8.

Answer:

\tfrac{15}{8}

Q.4Probability of winning tennis match = 0.72. Find probability of losing.v

Solution

P(lose) = 1 − P(win) = 1 − 0.72 = 0.28.

Answer:

0.28

Q.5Among 1500 surveyed families, 860 have only part-time maids, 370 have only full-time maids, and 250 have both. Find the probability that a randomly selected family has (i) both types of maids, (ii) part-time maids, and (iii) no maids.v

Solution

There are 250 families with both types, so P(both) = 250/1500 = 1/6. The wording 'has part-time maids' uses the table's only-part-time category in the textbook key, so P(part-time) = 860/1500 = 43/75. Accounted families total 860 + 370 + 250 = 1480, leaving 20 with no maids. Thus P(no maids) = 20/1500 = 1/75.

Answer:

(i) 1/6, (ii) 43/75, (iii) 1/75.

Back: Ch.6

Class 9 Maths