Answer and Step-by-step explanation:
First of all lets define
X to be that the bags do contain large liquid amounts
+ to be that the test is not negative that is positive
From the question,
P(x) = 3/100 = 0.03
P(+|X) = 0.91
P(+|X′) = 0.05
Probability of bag having a positive test =p(+)
= P(+|X) (P(x)) + P(+|X′)(P(X′))
= P(X′)) = 1 – 0.03 = 0.97
Inserting these values into these formulas
0.91)(0,03) + (0.05)(0.97)
= 0.0273 + 0.0485
= 0.0758
The probability of the randomly selected bag having large liqid amount
= P(X|+) = (0.91*0.03)/(0.91*0.03)(0.05*0.97)
= 0.0273/0.0273+0.0485
= 0.0273/0.0758
= 0.3602
the probability that this bag does not have large liquid amount
= p(X'|+) = 1 - P(X|+)
= 1 - 0.3602
= 0.6398
