Question 23 - CBSE Sample Paper Class 12 - 2017-18

Last updated at Dec. 21, 2017 by Teachoo

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.

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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 = /

CBSE Sample Paper Class 12 - 2017-18

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.