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Ex 15.1, 9 A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red? Total number of marbles = 5 + 8 + 4 = 17 Number of marbles which are red = 5 P(marble taken out is red) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘  π‘€β„Žπ‘–π‘β„Ž π‘Žπ‘Ÿπ‘’ π‘Ÿπ‘’π‘‘)/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘ ) = πŸ“/πŸπŸ• Ex 15.1, 9 A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (ii) white? Total number of marbles = 17 Number of marbles which are white = 8 P(marble taken out is white) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘  π‘€β„Žπ‘–π‘β„Ž π‘Žπ‘Ÿπ‘’ π‘€β„Žπ‘–π‘‘π‘’)/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘ ) = πŸ–/πŸπŸ• Ex 15.1, 9 (Method 1) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (iii) not green? P(marble taken out is not green) = P(marble taken out is white) + P(marble taken out is red) = 8/17 + 5/17 = πŸπŸ‘/πŸπŸ• Ex 15.1, 9 (Method 2) A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (iii) not green? P(marble taken out is not green) = 1 – P(marble taken out is green) Total number of marbles = 17 Number of marbles which are green = 4 P(marble taken out is green) = (π‘π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘  π‘€β„Žπ‘–π‘β„Ž π‘Žπ‘Ÿπ‘’ π‘”π‘Ÿπ‘’π‘’π‘›)/(π‘‡π‘œπ‘‘π‘Žπ‘™ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘šπ‘Žπ‘Ÿπ‘π‘™π‘’π‘ ) = πŸ’/πŸπŸ• P(marble taken out is not green) = 1 – P(marble taken out is green) = 1 – 4/17 = (17 βˆ’ 4)/17 = πŸπŸ‘/πŸπ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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.