Four bad oranges are accidentally mixed with 16 good ones. Find the probability distribution of the number of bad oranges when two oranges are drawn at random from this lot. Find the mean and variance of the distribution.

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

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Four bad oranges are accidentally mixed with 16 good ones. Find

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Question 23 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2

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Question 23 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3 Question 23 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4 Question 23 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5

  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Let X : be the number of bad oranges x 0 1 2 The Mean is given by : E( )" =" 2_( =1)^ = 0 60/95+1 32/95+2 3/95 = 38/95 = 2/5 = 0.4 Hence the mean is 0.4 The variance of x is given by : Var ( )= ( ^2 ) [ ( )]^2 So, finding ( ^2 ) E( ^ )= 2_( = 1)^ _ ^2 = 0^2 60/95+1^2 32/95+2^2 3/95 = 0 + 32/95 + 4 3/95 = 44/95 Now, Var ( )= ( ^ ) [ ( )]^ = 44/95 (2/5)^2 = 44/95 4/25 = 44/(19 5) 4/(5 5) = (44(5) 4 19)/(19 5 5) = (220 76)/475 = /

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.