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Ex 15.1, 17 - A lot of 20 bulbs contain 4 defective ones - Ex 15.1

  1. Chapter 15 Class 10 Probability
  2. Serial order wise
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Ex15.1,17 (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) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑏𝑢𝑙𝑏𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑢𝑙𝑏𝑠)   = 4/20 = 1/5 Ex15.1,17 (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 – 1 = 19 Number of defective bulbs = 4 Number of non-defective bulbs = 19 – 4 = 15 P (getting a non-defective bulb) = (𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑜𝑛− 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑏𝑢𝑙𝑏𝑠)/(𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑏𝑢𝑙𝑏𝑠)   = 15/19

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