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A factory produces cars that are safe 80% of the time and uses a machine to test whether each car is safe. The machine marks safe cars incorrectly 10% of the time and unsafe cars incorrectly 40% of the time. If a car is marked as safe by the machine, what is the probability that it is actually unsafe?

Respuesta :

Answer:

0.5

Explanation:

Given:

• P(Safe), P(S)=80%=0.8

,

• P(Unsafe), P(U) = 1-0.8=0.2

,

• P(Marked Incorrectly | Safe)=P(I|S)=10%=0.1

,

• P(Marked Incorrectly | Unsafe)=P(I|U)=40%=0.4

We want to find the probability that a car is marked incorrectly (safe) given that it is actually unsafe.

By Baye's formula for conditional probability:

[tex]\begin{gathered} P(Unsafe|Marked\text{ Incorrectly\rparen}=\frac{P(I|U)P(U)}{P(I|U)P(U)+P(I|S)P(S)} \\ =\frac{0.4\times0.2}{0.4\times0.2+0.1\times0.8} \\ =\frac{0.08}{0.08+0.08} \\ =\frac{0.08}{0.16} \\ =0.5 \end{gathered}[/tex]

The probability that a car marked as safe is actually unsafe is 0.5 (or 50%).

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