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The way you work this out is by thinking about the odds of each singular event, then finding the overall odds based on the individual odds.
The number of different books Mrs. Reid can choose is 9, so the first number is 9.
Mrs. Reid has picked one book so far, so now she has (1 - 9) = 8 books to choose from.
The of different books Mrs. Reid can choose now is 8, so your second number is 8.
Mrs. Reid has picked 2 book thus far, so now there are (2 - 9) = 7 books to choose from.
The of different books Mrs. Reid can choose now is 7, so your second number is 7.
To get the total number of different choices, multiply all the singular events together:
The number of different books Mrs. Reid can choose is 9, so the first number is 9.
Mrs. Reid has picked one book so far, so now she has (1 - 9) = 8 books to choose from.
The of different books Mrs. Reid can choose now is 8, so your second number is 8.
Mrs. Reid has picked 2 book thus far, so now there are (2 - 9) = 7 books to choose from.
The of different books Mrs. Reid can choose now is 7, so your second number is 7.
To get the total number of different choices, multiply all the singular events together:
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 .