{"id":44011,"date":"2022-05-29T11:30:18","date_gmt":"2022-05-29T06:00:18","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44011"},"modified":"2023-01-25T14:18:13","modified_gmt":"2023-01-25T08:48:13","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-19-ex-19-3","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-19-ex-19-3\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 19 Data Handling Ex 19.3"},"content":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 19 Data Handling Ex 19.3<\/h2>\n

Question 1.
\nList the outcomes you can see in these experiments.
\n\"ML
\nSolution:
\n(i) Outcomes in spinning wheel = A, A, A, B, C, D.
\n(ii) Outcomes in drawing a ball from a bag containing 5 identical balls
\nof different colours, says Red, black, green, blue, yellow.<\/p>\n

Question 2.
\nA die is rolled once. Find the probability of getting
\n(i) an even number
\n(ii) a multiple of 3
\n(iii) not a multiple of 3
\nSolution:
\nTotal outcomes of a die when rolled once:
\n1, 2, 3, 4, 5, 6 = 6
\n(i) Even numbers : 2, 4, 6
\ni.e. Favourable outcomes = 3
\n\u2234 Probability P(E) = \\(\\frac{3}{6}=\\frac{1}{2}\\)
\n(ii) Multiple of 3 = 3, 6
\ni.e. Favourable outcomes = 2
\n\u2234 Probability P(E) = \\(\\frac{2}{6}=\\frac{1}{3}\\)
\n(iii) Not a multiple of 3 = 1, 2, 4, 5
\ni.e., favourable outcomes = 4
\n\u2234 Probability P(E) = \\(\\frac{4}{6}=\\frac{2}{3}\\)<\/p>\n

Question 3.
\nTwo coins are tossed together. Find the probability of getting
\n(i) two tails
\n(ii) atleast one tail
\n(iii) no tail
\nSolution:
\nWhen two coins are tossed together, then
\nTotal outcomes = 2 \u00d7 2 = 4
\ni.e. HH, HT, TH, TT
\n(i) Favourable outcomes of getting two tails = 1
\n\u2234 Probability P(E) = \\(\\frac{1}{4}\\)
\n(ii) Favourable outcomes of getting atleast one tail
\nTH, HT, TT = 3
\n\u2234 Probability P(E) = \\(\\frac{3}{4}\\)
\n(iii) Favourable outcomes of getting No tail: HH = 1
\n\u2234 Probability P(E) = \\(\\frac{1}{4}\\)<\/p>\n

Question 4.
\nThree coins are tossed together. Find the probability of getting
\n(i) atleast two heads
\n(ii) atleast one tail
\n(iii) atmost one tail.
\nSolution:
\nThree coins are tossed together.
\n\u2234 Total outcomes = 8
\n= HHH, HHT, HTH, THH, HTT, TTH, TTT, THT
\n(i) Favourable outcomes of getting
\nAtleast two heads : These can be HHH, HHT, HTH, THH = 4 in numbers
\n\u2234 P(E) = \\(\\frac{\\text { Number of favourable outcome }}{\\text { Number of all possible outcome }}=\\frac{4}{8}=\\frac{1}{2}\\)
\n(ii) Favourable outcomes of getting
\nAtleast one tail: There can be HHT, HTH, HTT, TTT, THH, THT, TTH = 7 in number
\n\u2234 P(E) = \\(\\frac{\\text { Number of favourable outcome }}{\\text { Number of all possible outcome }}=\\frac{7}{8}\\)
\n(iii) Favourable outcomes of getting
\nAtmost one tail: There can be HHH, HHT, HTH, THH = 4
\n\u2234 P(E) = \\(\\frac{\\text { Number of favourable outcome }}{\\text { Number of all possible outcome }}=\\frac{4}{8}=\\frac{1}{2}\\)<\/p>\n

Question 5.
\nTwo dice are rolled simultaneously. Find the probability of getting
\n(i) the sum as 7
\n(ii) the sum as 3 or 4
\n(iii) prime numbers on both the dice.
\nSolution:
\nTwo dice are rolled simultaneously
\nTotal outcomes = 6 \u00d7 6 = 36
\n(i) Sum as 7 : (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) = 6
\n\u2234 Probability P(E) = \\(\\frac{\\text { Favourable oucomes }}{\\text { Total outcomes }}=\\frac{6}{36}=\\frac{1}{6}\\)
\n(ii) The sum as 3 or 4
\n(1, 2), (1, 3), (2, 1), (2, 2), (3, 1) = 5
\n\u2234 Probability P(E) = \\(\\frac{\\text { Favourable oucomes }}{\\text { Total outcomes }}=\\frac{5}{36}\\)
\n(iii) Prime numbers on both the dice
\n(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5) = 9
\n\u2234 Probability P(E) = \\(\\frac{\\text { Favourable oucomes }}{\\text { Total outcomes }}=\\frac{9}{36}=\\frac{1}{4}\\)<\/p>\n

Question 6.
\nA fcox contains 600 screws, one tenth are rusted. One screw is taken out at random from the box. Find the probability that it is
\n(i) a rusted screw
\n(ii) not a rusted screw
\nSolution:
\nRusted screw = \\(\\frac{1}{10}\\) of 60\u00b0 = \\(\\frac{1}{10}\\) \u00d7 600 = 60 seconds
\n(i) Favourable outcomes of picking rusted screw = 60
\nP(E) = \\(\\frac{60}{600}=\\frac{1}{10}\\)
\n(ii) P (of not a rusted screw) = 1 – P (rusted screw) = 1 – \\(\\frac{1}{10}=\\frac{9}{10}\\)<\/p>\n

Question 7.
\nA letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel?
\nSolution:
\n‘TRIANGLE’
\nTotal number of outcomes = 8
\nVowels : I, A, E = 3
\n\u2234 Probability P(E) = \\(\\frac{3}{8}\\)<\/p>\n

Question 8.
\nA bag contains 5 red, 6 black and 4 white balls. A ball is drawn at random from the bag, find the probability the ball is drawn is
\n(i) white
\n(ii) not black
\n(iii) red or black
\n(iv) neither red nor black
\nSolution:
\nIn a bag, there are 5 red, 6 black and 4 white balls.
\nTotal number of outcomes = 5 + 6 + 4=15
\n(i) Probability of white ball = \\(\\frac{4}{15}\\)
\n(ii) Probability of not black (5 + 4 = 9) balls = \\(\\frac{9}{15}=\\frac{3}{5}\\)
\n(iii) Probability of red or black ball (5 + 6 = 11)= \\(\\frac{11}{15}\\)
\n(iv) Probability of ball which is neither red nor black, white ball (4) P(E) = \\(\\frac{4}{15}\\)<\/p>\n

Question 9.
\nA box contains 17 cards numbered 1, 2, 3, ……….,17 and are mixed thoroughly. A card is drawn at random from the box. Find the probability that the number on the card is
\n(i) odd
\n(ii) even
\n(iii) prime
\n(iv) divisible by 3
\n(v) divisible by 2 and 3 both
\nSolution:
\nA box contains 17 cards numbered 1 to 17
\nTotal number of outcomes = 17
\n(i) Card bearing odd number (1, 3, 5, 7, 9, 11, 13, 15, 17) = 9
\n\u2234Probability P(E) = \\(\\frac{9}{17}\\)
\n(ii) Even number (2, 4, 6, 8, 10, 12, 14, 16) = 8
\n\u2234Probability P(E) = \\(\\frac{8}{17}\\)
\n(iii) Prime numbers {2, 3, 5, 7, 11, 13, 17) = 7
\n\u2234Probability P(E) = \\(\\frac{7}{17}\\)
\n(iv) Numbers divisible by 3 = 3, 6, 9, 12, 15 = 5
\n\u2234Probability P(E) = \\(\\frac{5}{17}\\)
\n(v) Number divisible by 2 and 3 both 6, 12 = 2
\n\u2234Probability P(E) = \\(\\frac{2}{17}\\)<\/p>\n

Question 10.
\nA card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is:
\n(i) an ace
\n(ii) a red card
\n(iii) neither a king nor a queen
\n(iv) a red face card or an ace
\n(v) a card of spade
\n(vi) non-face card of red colour.
\nSolution:
\nNumber of playing cards in a deck = 52
\nOne card is drawn
\n(i) An ace : 4
\n\u2234 Probability P(E) = \\(\\frac{4}{52}=\\frac{1}{13}\\)
\n(ii) A red card = 13 + 13 = 26
\n\u2234 Probability P(E) = \\(\\frac{26}{52}=\\frac{1}{2}\\)
\n(iii) Neither a king nor a queen
\nNumber of cards 52 – (4 + 4) = 52 – 8 = 44
\n\u2234 Probability P(E) = \\(\\frac{44}{52}=\\frac{11}{13}\\)
\n(iv) A red face card = 6
\n\u2234 Probability P(E) = \\(\\frac{6}{52}=\\frac{3}{26}\\)
\n(v) A card of spade or an ace = 13 + 3 = 16
\n\u2234 Probability P(E) = \\(\\frac{16}{52}=\\frac{4}{13}\\)
\n(vi) Non-face card of red colour = 26 – 6 = 20
\n\u2234 Probability P(E) = \\(\\frac{20}{52}=\\frac{5}{13}\\)<\/p>\n

Question 11.
\nIn a lottery, there are 5 prized tickets and 995 blank tickets. A person buys a lottery ticket. Find the probability of his winning a prize.
\nSolution:
\nNumber of prized tickets = 5
\nNumber of blank tickets = 995
\nTotal number of tickets = 5 + 995 = 1000
\nProbability of prized ticket
\n\\(P(E)=\\frac{\\text { Number of favourable outcome }}{\\text { Number of possible outcome }}\\)
\n= \\(\\frac{5}{1000}=\\frac{1}{200}\\)<\/p>\n

ML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 19 Data Handling Ex 19.3 Question 1. List the outcomes you can see in these experiments. Solution: (i) Outcomes in spinning wheel = A, A, A, B, C, D. (ii) Outcomes in drawing a ball from a bag containing 5 identical balls of different colours, says … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\nML Aggarwal Class 8 Solutions for ICSE Maths Chapter 19 Data Handling Ex 19.3 - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-19-ex-19-3\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 19 Data Handling Ex 19.3\" \/>\n<meta property=\"og:description\" content=\"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 19 Data Handling Ex 19.3 Question 1. 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List the outcomes you can see in these experiments. Solution: (i) Outcomes in spinning wheel = A, A, A, B, C, D. (ii) Outcomes in drawing a ball from a bag containing 5 identical balls of different colours, says ... 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