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) Find the probability that a failure is due to loose keys. b) Find the probability that a failure is due to improperly connected or poorly welded wires.

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 = 0.078Step-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 [tex]\bigcup[/tex] PWW) ;  P(IC [tex]\bigcup[/tex] PWW) = P(IC) + P(PWW) - P(IC [tex]\bigcap[/tex] PWW) { Here P(IC [tex]\bigcap[/tex] 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 [tex]\bigcup[/tex] PWW) = 0.0156 + 0.0676 - 0 = 0.078 .