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Sunday, December 25, 2016

bayes theorem problem 2 and 3


A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Let E1 be the event that the first bag is selected.

let E2 be the event that the second bag is selected.

let A be the event of drawing a red ball from the selected bag.

We can assume that E1 and E2 are equally likely so that

P( E1 ) = 1 / 2

P( E2 ) = 1 / 2

P( A / E1 ) = 4 / ( 4 + 4 )  = 4 / 8 [ 4 red balls out of a total of 4 + 4 = 8 balls in the first bag ]

P( A / E2 ) = 2 / ( 2 + 6 ) = 2 / 8  [ 2 red balls out of a total of 2 + 6 = 8 balls  in the second bag]

By bayes theorem ,

 probability that the ball is drawn from the first bag = P ( E1 / A )

                         
               



P ( E1 / A ) =  { (1/2)(4/8) } / { (1/2)(4/8) + (1/2)(2/8) }


P ( E1 / A ) = 4 / 6 =  2 / 3

index of more problems on baye's theorem for ncert cbse mathematics


problem 3

Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostelier?


Let E1 be the event that the chosen student is a hostelier.

let E2 be the event that the chosen student is a day scholar.

let A be the event that the chosen student has a A grade in the annual exam.

P( E1 ) = 60 / 100

P( E2 ) = 40 / 100

P( A / E1 ) = 30 / 100 [ Previous year results report that 30% of all students who reside in hostel attain A grade ]

P( A / E2 ) = 20 / 100  [ Previous year results report that 20% of day scholars attain A grade  ]

Required probability = P [ chosen student is a hostelier given that he has an A grade ]

Required probability = P [ E1 / A ]

  



P ( E1 / A ) = [ (60 / 100 ) ( 30 / 100 )] / { [ (60 / 100 ) ( 30 / 100 ) ] + [ (40 / 100 ) ( 20 / 100 ) ] }


P ( E1 / A ) = ( 18 / { 18 + 8 } ) =  ( 18 / 26 ) = ( 9 / 13 )

index of more problems on baye's theorem for ncert cbse mathematics

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bayes theorem is a topic for cbse / ncert  / scert of 12th standard in India. These are some of the important questions and their solutions from the topic of bayes theorem from various cbse ncert textbooks and old question papers.

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