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

(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

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 .