RS Aggarwal Solutions for Class 10 Maths Chapter 19Probability

Exercise 19A Question 1: Solution: (i) The probability of an impossible event is zero. (ii) The probability of a sure event is one. (iii) For any event E, P(E) + P(not E) = 1. (iv) The probability of a possible but not a sure event lies between zero and one. (v) The sum of probabilities of all the outcomes of an experiment is one. Question 2: Solution: A coin is tossed: Sample space = {H, T} Total number of outcomes = 2 P(getting a tail) = 1/2. Question 3: Solution: Two coins are tossed: Sample space = {HH, HT, TH, TT} Total number of outcomes = 4 P(getting a tail) = 1/2. Now, (i) P(exactly 1 head) = 2/4 = 1/2 (ii) P(at most 1 head) = 3/4 (iii) P(at least 1 head) = 3/4 Question 4: Solution: In a throw of a dice, Possible outcomes = {1, 2, 3, 4, 5, 6} Total number of possible outcomes = 6 Now, (i) Favorable outcomes = Even numbers = {2, 4, 6} Number of favorable outcomes = 3 P(an even number) = 3/6 = 1/2. (ii) Favorable outcomes = number less than 5 = { 1, 2, 3, 4} Number of favorable outcomes = 4 P(a number less than 5) = 4/6 = 2/3. (iii) Favorable outcomes = number greater than 2 ={ 3, 4, 5, 6} Number of favorable outcomes = 4 P(a number greater than 2) = 4/6 = 2/3. (iv)

RS Aggarwal Solutions for Class 10 Maths Chapter 19Probability Favorable outcomes = a number between 3 and 6 = { 4, 5} Number of favorable outcomes = 2 P(a number between 3 and 6) = 2/6 = 1/3. (v) Favorable outcomes = a number other than 3 = {1, 2, 4, 5, 6} Number of favorable outcomes = 5 P(a number other than 3) = 5/6. (vi) Favorable outcome = a number 5 = { 5} Number of favorable outcomes = 1 P(a number 5) = 1/6. Question 5: Solution: Total numbers of alphabet = 26 Total numbers of consonants = 21 P(choosing a consonant) = 21/26 Question 6: Solution: Total letter on the dice = 6 (i) Number of times A appears = 3 Therefore, numbers of favorable outcomes = 3 P (getting letter A) = 3/6 = 1/2 (ii) Number of times D appears = 1 Therefore, numbers of favorable outcomes = 1 P (getting letter D) = 1/6 Question 7: Solution: Given: Total number of bulbs = 200 Number of defective bulbs = 16 (i)Let A be the event of getting a defective bulb Total number of defective bulbs = 16 So, P(getting defective bulbs) = P(A) = 16/200 = 2/25 (ii)Let B be the event of getting non-defective bulb So, P(getting non-defective bulb) = P(B) = 1 – P(A) = 1 – 2/25 = 23/25. Question 8: Solution: Probability of winning a game = 0.7 Let us say, P(A) = 0.7 Let B be probability of losing the game. Find: P(B) We know that, P(A) + P(B) = 1 => P(B) = 1-0.7 = 0.3

RS Aggarwal Solutions for Class 10 Maths Chapter 19Probability Question 9: Solution: Total numbers of students = 35 Number of girls = 15 Number of boys = 20 (i) Numbers of favorable outcome = 20 P(choosing a boy) = 20/35 = 4/7 (ii) Numbers of favorable outcome are= 15 P(choosing a girl) = 15/35 = 3/7 Question 10: Solution: Total numbers of lottery tickets = 10 + 25 = 35 P(getting a prize) = 10/35 = 2/7 Question 11: Solution: Total number of tickets sold = 250 Number of prizes = 5 Number of favorable outcomes = 5 P(getting a prize) = 5/250 = 1/50 Question 12: Solution: Total numbers of outcomes = 17 (i) Odd numbers on the cards = { 1, 3, 5, 7, 9, 11, 13, 15, 17 } Numbers of favorable odd numbers = 9 P (getting an old number) = 9/17 (ii) Numbers divisible by 5 form 1 to 17 = {5, 10, 15} Number of favorable outcome = 3 P (getting a number divisible by 5) = 3/17 Question 13: Solution: Total number of outcomes = 8 Factors of 8 from 1 to 8 are {1,2,4,8} Total number of favorable outcomes = 4 Therefore, P(getting a factor of 8 ) = 4/8 = ½ Question 14: Solution: Let E be the event of having at least one boy The probability that each child will be a girl= 1/2 The probability that each child will be a boy = 1/2 The probability of no boys = All are girls = 1/2 x 1/2 x 1/2 = 1/8

RS Aggarwal Solutions for Class 10 Maths Chapter 19Probability Then, the number of favorable outcome is P(E) –P(not E)=1 where Let E be the event of having at least one boy, then P(E) is probability of having atleast one boy and P(not E) = Probability of having no boys. Therefore, Probability (at least 1 boy) = 1 – Probability (no boys)= 1 – 1/8 = 7/8 Question 15: Solution: Probability = (number of favorable outcomes)/(Total number of outcomes) A bag contains 4 white balls, 5 red balls, 2 black balls and 4 green balls. Total numbers of balls = 4 + 5 + 2 + 4 = 15 (i) Total numbers of black balls = Numbers of favorable outcomes= 2 P(getting a black ball) = 2/15 (ii) Total numbers of green balls = 4 Numbers of non-green balls = 15 – 4 = 11 P(not green ball) = 11/15 (iii) Total numbers of red and white balls = 5 + 4 = 9 P(red ball or white ball) = 9/15 = 3/5 (iv) Total numbers of red and green balls = 5 + 4 = 9 Number of balls which are neither red nor green = 15 – 9 = 6 P (getting neither red nor green ball) = 6/15 = 2/5 Question 16: Solution: Total number of cards in a well-shuffled pack=52 (i) Number of red king cards = 2 P(a red king) = 2/52 = 1/26 (ii) Number of queen cards = 4 Number of jack cards = 4 Total number of favorable outcomes = 4 + 4 = 8 P(a queen or a jack) = 8/52 = 2/13 Question 17: Solution: Total number of cards in a well-shuffled pack = 52 Total number of queens and kings cards = 4 + 4 = 8 Therefore, there are 44 cards (52 – 8) that are neither king nor queen. Total number of favorable outcomes = 44 Required probability = 44/52 = 11/13 Question 18:

RS Aggarwal Solutions for Class 10 Maths Chapter 19Probability Solution: Total number of cards in a well-shuffled pack = 52 (i) Number of red face cards = 6 P (a red face card) = 6/52 = 3/26 (ii) Number of black king cards = 2 P (a black king) = 2/52 = 1/26 Question 19: Solution: The possible outcomes if two dice are tossed together are

Total number of outcomes = 36 (i) The favorable outcomes = getting an even number on each die = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)} The number of favorable outcomes = 9 P(getting even number on both dice) = 9/36 = 1/4 (ii) The favorable outcomes = sum of the numbers appearing on two dice is 5 = {(1, 4), (2, 3), (3, 2), (4, 1)} The numbers of favorable outcomes = 4 P(sum of numbers appearing on two dice is 5) = 4/36 = 1/9 Question 20: Solution: When two different dice are rolled simultaneously, then Total number of outcomes = 36 Favorable outcomes = Sum of the numbers on the two dice is 10 = {(2, 5),(5,2), (5,5)} Total number of favorable outcomes = 3 P(getting numbers on the dice whose sum is 10) = 3/36 = 1/12