Question

a building contractor buys 80% of his cement from supplier a and 20% from supplier b. a total of 80% of the bags from a arrive undamaged, while 75% of the bags from b arrive undamaged. find the probability that an undamaged bag is from supplier a.
the probability is
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Define the probabilities

Let $P(A)=0.8$ (probability of buying from supplier A), $P(B) = 0.2$ (probability of buying from supplier B), $P(U|A)=0.8$ (probability of a bag being undamaged given it's from supplier A), $P(U|B)=0.75$ (probability of a bag being undamaged given it's from supplier B).

Step2: Calculate the total probability of a bag being undamaged

By the law of total - probability, $P(U)=P(U|A)P(A)+P(U|B)P(B)$.
$P(U)=(0.8\times0.8)+(0.75\times0.2)=0.64 + 0.15=0.79$.

Step3: Use Bayes' theorem

We want to find $P(A|U)$. By Bayes' theorem, $P(A|U)=\frac{P(U|A)P(A)}{P(U)}$.
Substitute the known values: $P(A|U)=\frac{0.8\times0.8}{0.79}=\frac{0.64}{0.79}\approx0.8101$.

Answer:

$0.8101$