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.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


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

Question 23 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 23 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3
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Question 23 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5


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 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.