1 over 7 to the 3 power

Consider this:

If a test gives a positive result for an infected person 98% of the time, that means that 2% of the time, it gives a negative result for an infected person, which would be a false negative.

Then,

if the test is 97& accurate (not precise) for non-infected people, that means that it gives a negative result 97& of the time. So a positive result will be given 3% of the time for non-infected people, which is considered a false positive.

5 0

3 years ago

Read 2 more answers

Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are relat

anzhelika [568]

Answer:

(c) Probability that a failure is due to loose keys = 0.2376

(d) Probability that a failure is due to improperly connected or poorly welded wires = 0.078

Step-by-step explanation:

The Whole probability scenario is given for Computer Keyboard failures.

(a) Let F be the event of failure due to faulty electrical connects, P(F) = 0.12

M be the event of failure due to mechanical defects, P(M) = 0.88

LK be the event of mechanical defect due to loose keys, P(LK/M) = 0.27

IA be the event of mechanical defect due to improper assembly, P(IA/M) =0.73

DW be the event of electrical connects due to defective wires,P(DW/F) = 0.35

IC be the event of electrical connects due to improper connections,

P(IC/F) = 0.13 .

PWW be the event of electrical connects due to poorly welded wires,

P(PWW/F) = 0.52

(b) Keyboard failures

Faulty electrical connects Mechanical Defects

P(F) = 0.12 P(M) = 0.88

Defective wires Improper Poorly Loose Keys Improper

P(DW/F)=0.35 Connections Welded wires P(LK/M)=0.27 Assembly

P(IC/F)=0.13 P(PWW/F)=0.52 P(IA/M)=0.73

This is the required tree diagram.

(c) Probability that a failure is due to loose keys is given by:

P(LK) =P(LK/M) * P(M) {This means mechanical failure is due to loose

keys}

P(LK) = 0.27 * 0.88 = 0.2376 .

(d) Probability that a failure is due to improperly connected or poorly welded

wires is given by P(IC $\bigcup$ PWW) ;

P(IC $\bigcup$ PWW) = P(IC) + P(PWW) - P(IC $\bigcap$ PWW) { Here P(IC $\bigcap$ PWW) = 0 }

P(IC) = P(IC/F) * P(F) = 0.13 * 0.12 = 0.0156

P(PWW) = P(PWW/F) * P(F) = 0.52 * 0.13 = 0.0676

Therefore, P(IC $\bigcup$ PWW) = 0.0156 + 0.0676 - 0 = 0.078 .

8 0

3 years ago