The Transportation Safety Authority (TSA) has developed a new test to detect large amounts of liquid in luggage bags. Based on many test runs, the TSA determines that if a bag does contain large amounts of liquid, there is a probability of 0.91 the test will detect it.
If a bag does not contain large amounts of liquid, there is a 0.05 probability the test will conclude that it does (a false positive).
Suppose that in reality only 3 in 100 bags actually contain large amounts of liquid.

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

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1