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)                                     <u> </u><u>Keyboard failures</u>
<h2>                              /               \</h2>
            <u> </u><u> Faulty electrical connects   </u>            <u>Mechanical Defects </u>           
                       P(F) = 0.12                                             P(M) = 0.88
<h2>        /            |             \                  /            \</h2>
<u><em>Defective wires</em></u>  <u><em>Improper</em></u>        <u><em>Poorly</em></u>                  <u><em>Loose Keys</em></u>      <u><em>Improper</em></u><em> </em>
 P(DW/F)=0.35   <u><em>Connections</em></u>   <u><em>Welded wires</em></u>      P(LK/M)=0.27   <em> </em><u><em>Assembly</em></u>
                            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  PWW) ;
 PWW) ;
  P(IC  PWW) = P(IC) + P(PWW) - P(IC
 PWW) = P(IC) + P(PWW) - P(IC  PWW) { Here P(IC
 PWW) { Here P(IC  PWW) = 0 }
 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  PWW) = 0.0156 + 0.0676 - 0 = 0.078 .
 PWW) = 0.0156 + 0.0676 - 0 = 0.078 .