Question

Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are related to loose keys (27%) or improper assembly (73%). Electrical connect defects are caused by defective wires (35%), improper connections (13%), or poorly welded wires (52%).A) Define all relevant elemental events, and express all numbers in terms of the appropriate probabilities.B) Draw a tree diagram to illustrate the possible outcomes of the experiment of randomly selecting a keyboard failure and observing its source and cause.C) Find the probability that a failure is due to loose keys.D) Find the probability that a failure is due to improperly connected or poorly welded wires.

Answer 1

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 .