miscellaneous problem on bayes theorem 14 and 15

miscellaneous problem
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of a certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Let E1 be the event that the patient follows a course of meditation and yoga.

let E2 be the event that the patient follows the prescription of a certain drug.

let A be the event that the patient sufferes a heart attack.


P( E1 ) = ( 1 / 2 )  { given a patient can choose any one of the two options with equal probabilities}

P( E2 ) = ( 1 / 2 )  { given a patient can choose any one of the two options with equal probabilities}

P( A / E1 ) = ( 70 / 100 )( 40 / 100 ) { If E1 occurs, risk of heart attack is less by 30% }

P( A / E2 ) = ( 75 / 100 )( 40 / 100 ) { If E2 occurs, risk of heart attack is less by 25% }


Required probability = P [ the patient follows a course of meditation and yoga given that the patient sufferes a heart attack ]

Required probability = P [ E1 / A ]



P(E1 / A)=[( 1 / 2 )( 70 / 100 )( 40 / 100 ) ] / {[(1 / 2)(70/100)(40 / 100)] + [(1 / 2)(75/100)(40 / 100)]  }


P ( E1 / A ) =  [70] / { [70] + [75] } = ( 14 / 29 )
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problem 15

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls.One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Let E1 be the event that the ball transferred from Bag I to Bag II is black.

let E2 be the event that the ball transferred from Bag I to Bag II is red.

let A be the event that the ball drawn from Bag II after the transfer is red.


P( E1 ) = ( 4 / 7)  { 4 black in Bag I out of a total of 3 + 4 = 7  }

P( E2 ) = ( 3 / 7 )  { 3 red in Bag I out of a total of 3 + 4 = 7}

P( A / E1 ) = ( 4 / 10 ) { If E1 occurs, 1 more black makes 4R and 6B in Bag II }

P( A / E2 ) = ( 5 / 10 ) { If E2 occurs, 1 more red makes 5R and 5B in Bag II }


Required probability = P [ transferred ball is black given that ball drawn from Bag II after the transfer is red ]

Required probability = P [ E1 / A ]



P(E1 / A)=[( 4 / 7 )( 4 / 10 ) ] / {[( 4 / 7 )( 4 / 10 ) ] + [( 3 / 7 )( 5 / 10 ) ]  }


P ( E1 / A ) =  [16] / { [16] + [15] } = ( 16 / 31 )

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