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Ex  15.1, 2 - 1500 families with 2 children were selected - Ex 15.1

  1. Chapter 15 Class 9 Probability
  2. Serial order wise
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Ex 15.1, 2 1500 families with 2 children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having 2 girls Total number of families = 475 + 814 + 211 = 1500 Number of families having 2 girls = 475 P (a randomly chosen family has 2 girls) = 475/1500 = 19/60 Ex 15.1, 2 Compute the probability of a family, chosen at random, having (ii) 1 girl Total number of families = 1500 Number of families having 1 girl = 814 P (a randomly chosen family has 1 girl) = 814/1500 = 507/750 Ex 15.1, 2 Compute the probability of a family, chosen at random, having (iii) No girl Total number of families = 1500 Number of families having no girl = 211 P (a randomly chosen family has no girl) = 211/1500 Ex 15.1, 2 Also check whether the sum of these probabilities is 1. Sum of all these probabilities = 19/60 + 407/750 + 211/1500 = 475/1500 + 814/1500 + 211/1500 = 1500/1500 = 1 Hence, sum of probabilities is 1

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
  • Drrajendra Singh's image
    Drrajendra Singh
    June 12, 2016, 4:52 p.m.

    Sir , if I learn and remember the key log and anti log values than....

    1- how can I find out log and anti log approximate value on the basis of mean values ex if I know log  4 and log 6 value than I can easygoing find value for log 5 it will be intermediate of both . Means for time saving immidiate approach answer for anti log how can we?.

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