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Ex 14.1, 13 A die is thrown once. Find the probability of getting (i) a prime number; Total outcomes that can occur are 1, 2, 3, 4, 5, 6 Number of possible outcomes of a dice = 6 Prime number is a number not divisible by any number except itself Prime numbers on a dice are 2, 3, and 5. Total prime numbers on a dice = 3 Probability of getting a prime number = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘  ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘๐‘œ๐‘š๐‘’๐‘ )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘ ) = 3/6 = ๐Ÿ/๐Ÿ Ex 14.1, 13 A die is thrown once. Find the probability of getting (ii) a number lying between 2 and 6; Total outcomes that can occur are 1, 2, 3, 4, 5, 6 Number of possible outcomes of a dice = 6 Numbers lying between 2 and 6 = 3, 4, 5 Total numbers lying between 2 and 6 = 3 Probability of getting a number lying between 2 & 6 = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘  ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘–๐‘  ๐‘Ž ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘› 2 & 6)/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘ ) = 3/6 = ๐Ÿ/๐Ÿ Ex 14.1, 13 A die is thrown once. Find the probability of getting (iii) an odd number. Total outcomes that can occur are 1, 2, 3, 4, 5, 6 Number of possible outcomes of a dice = 6 Numbers which are odd = 1,3, 5 Total numbers which are odd = 3 Probability of getting an odd number = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘  ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘กโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘–๐‘  ๐‘Ž๐‘› ๐‘œ๐‘‘๐‘‘ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ)/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘œ๐‘ข๐‘ก๐‘๐‘œ๐‘š๐‘’๐‘ ) = 3/6 = ๐Ÿ/๐Ÿ

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo