Question

Airbags are manufactured by Aces (A), Best (B), and Cool (C) at rates of 57%, 26% and 17%, respectively. Airbags occasionally kill (K) passengers when they deploy in accidents. Airbags made by Aces, Best, and Cool do not kill people at rates of 99%, 96%, and 87%, respectively. One airbag is randomly selected for testing.

If an airbag kills a passenger, calculate the probability that the airbag was manufactured by Cool. (Round to the nearest ten-thousandth.)

Answer 1

Air bag manufactured by Aces(A)=57%

So, Probability $P(A)=\frac{57}{100}$

Air bag manufactured by Best (B) =26%

So, Probability $P(B)=\frac{26}{100}$

Airbag manufactured by Cool(C)=17%

So, Probability $P(C)=\frac{17}{100}$

Airbags made by Aces, Best, and Cool do not kill people at rates of 99%, 96%, and 87%, respectively.

Let K be the event which kill people.

Probability of Air bag made by A which kill people $P(K/A)=\frac{1}{100}$

Probability of Air bag made by B which kill people $P(K/B)=\frac{4}{100}$

Probability of Air bag made by C which kill people $P(K/A)=\frac{13}{100}$

If an airbag kills a passenger, calculate the probability that the airbag was manufactured by Cool

Using Baye's theorem:

$P(C/K)=\frac{P(K/C)P(C)}{P(K/A)P(A)+P(K/B)P(B)+P(K/C)P(C)}$

Substitute the values of probabilities into formula

We get,

$P(C/K)=\frac{0.17\times 0.13}{0.57\times 0.01+0.26\times 0.04+0.17\times 0.13}$

Now we calculate it and get probability

So, $P(C/K)=0.5785$

So, 57.85% of passenger kills if the airbag was manufactured by Cool.