Respuesta :
The probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition is
P( A | B ) = P( B | A ) P ( A ) / P( B ) so P ( Disease | positive ) = P ( positive |
disease ) P ( disease ) / P ( positive ) = (0.99)(0.03) / P ( positive )
P ( Positive ) = P( Positive | disease ) P ( disease )+ P( Positive | no disease
) P( no disease )
= (0.99)(0.03) + (0.02)(0.97) = 0.0491
so P( Disease | positive ) = (0.99)(0.03) / (0.0491) = .6
P( A | B ) = P( B | A ) P ( A ) / P( B ) so P ( Disease | positive ) = P ( positive |
disease ) P ( disease ) / P ( positive ) = (0.99)(0.03) / P ( positive )
P ( Positive ) = P( Positive | disease ) P ( disease )+ P( Positive | no disease
) P( no disease )
= (0.99)(0.03) + (0.02)(0.97) = 0.0491
so P( Disease | positive ) = (0.99)(0.03) / (0.0491) = .6
The probability that a randomly selected person that tests positive for the predisposition by the test actually has the predisposition is : 0.6
i.e. P( Disease | positive ) = 0.6
Let : B = positive , A = disease
∴ P( A | B ) = P( B | A ) P ( A ) / P( B )
where : P( B | A ) = 0.99
P ( A ) = 0.03
∴ P(Disease | positive ) = P ( positive | disease ) * P( disease ) / P ( positive )
= (0.99) * (0.03) / P(positive ) --------- ( 1 )
Also P( Positive ) = P( B | A ) P( A )+ P( B | no disease
)* P( no disease )
= (0.99)*(0.03) + (0.02)* (0.97) = 0.049 ------- ( 2 )
back to equation ( 1 )
P( Disease | positive ) = (0.99)* (0.03) / (0.049) = 0.6.
Hence the probability that a randomly selected person person who tests positive has the predisposition is 0.6
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