problem 6 and problem 7 of bayes theorem

There are three coins. One is a two headed coin (having head on both faces),another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?


Let E1 be the event that the selected coin is the two headed coin .

let E2 be the event that the selected coin is the biased coin that comes up heads 75% of the time.

let E3 be the event that the selected coin is the unbiased coin.

let A be the event that the  toss of the selected coin resulted in a head.

Assuming E1 , E2 , E3 are equally likely

P( E1 ) = ( 1 / 3 )

P( E2 ) = ( 1 / 3 )

P( E3 ) = ( 1 / 3 )

P( A / E1 ) = 1 { since the coin is two headed  in the event of E1 }

P( A / E2 ) = ( 75 / 100 ) = ( 3/4 )  { since the biased coin comes up heads 75% of the time in E2 }

P( A / E3 ) = ( 1 / 2 )  { since the  coin is unbiased in E3 }

Required probability = P [ a person that the two headed coin was selected given that the toss resulted in a head ]

Required probability = P [ E1 / A ]


              



P ( E1 / A ) = [ ( 1 / 3 )( 1 )] / { [( 1 / 3 )( 1 )] + [( 1 / 3 )( 3/4 ) ] + [( 1 / 3 )( 1/2 ) ] }


P ( E1 / A ) =  [1] / { [1]+[3/4]+[1/2] } =  ( 4 /9 )

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

problem 7

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Let E1 be the event that the person is a scooter driver .

let E2 be the event that  the person is a car driver.

let E3 be the event that the person is a truck driver.

let A be the event that the  person met with an accident.

total number of vehicles = 2000 + 4000 + 6000 = 12000

P( E1 ) = ( 2000 / 12000 )

P( E2 ) = ( 4000 / 12000 )

P( E3 ) = ( 6000 / 12000 )

P( A / E1 ) = 0.01

P( A / E2 ) = 0.03

P( A / E3 ) = 0.15

Required probability = P [ person is a scooter driver given that the person met with an accident ]

Required probability = P [ E1 / A ]


         


P ( E1 / A ) = [(2000 / 12000 ) ( 0.01 )] / { [(2000 / 12000) (0.01)] + [(4000 / 12000) (0.03)] + [(6000 / 12000) (0.15)] }


P ( E1 / A ) =  [20] / { [20]+[120]+[900] } =  ( 20 / 1040 ) = ( 1 / 52 )

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

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