Eighteen students are members of the Science Club. This is 1/3 the number of students in the three sections of
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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) ;
P(IC PWW) = P(IC) + P(PWW) - P(IC
PWW) { Here P(IC
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 .