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Ex 14.1, 17 (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Total number of bulbs = 20 Total number of defective bulbs = 4 P (getting a defective bulb) = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘‘๐‘’๐‘“๐‘’๐‘๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘๐‘ข๐‘™๐‘๐‘ )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘ข๐‘™๐‘๐‘ ) = 4/20 = ๐Ÿ/๐Ÿ“ Ex 14.1, 17 (ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective? Total number of bulbs left = 20 โ€“ 1 = 19 Number of defective bulbs = 4 Number of non-defective bulbs = 19 โ€“ 4 = 15 P (getting a non-defective bulb) = (๐‘๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘›๐‘œ๐‘›โˆ’ ๐‘‘๐‘’๐‘“๐‘’๐‘๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘๐‘ข๐‘™๐‘๐‘ )/(๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘๐‘ข๐‘™๐‘๐‘ ) = ๐Ÿ๐Ÿ“/๐Ÿ๐Ÿ—

  1. Chapter 14 Class 10 Probability
  2. Serial order wise

About the Author

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