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Answer:
The probability that a 10-year old computer still has fully functioning MB and HD is 45%
Step-by-step explanation:
The probability of the union of two events, is equal to the sum of the individual probabilities of the two events, minus the probability of the intersection event.
That means that the probability of have a fully functional computer (PFC) is equal to the sum of probability of have a computer with a functional motherboards (PFMB) plus the probability of have a computer with a functional hard drives (PFHD), minus the probability of have a computer with both motherboards and hard drives functional (PBMH).
PFC = PFMB+PFHD-PBMH
- Probability of functional motherboards (PFMB)
PFMB = 1 - probability of having problems with motherboards = 1 - 40 %
PFMB = 1 - [tex]\frac{40}{100}[/tex] = 1 - [tex]\frac{4}{10}[/tex] = [tex]\frac{6}{10}[/tex]
- Probability of functional hard drive (PFHD)
PFHD = 1 - probability of having problems with hard drive = 1 - 30 %
PFHD = 1 - [tex]\frac{30}{100}[/tex] = 1 - [tex]\frac{3}{10}[/tex] = [tex]\frac{7}{10}[/tex]
- Probability of functional both (PFMH)
PFMH = 1 - probability of having problems with both = 1 - 15 %
PFMH = 1 - [tex]\frac{15}{100}[/tex] = 1 - [tex]\frac{3}{20}[/tex] = [tex]\frac{17}{20}[/tex]
PFC = [tex]\frac{6}{10}[/tex] + [tex]\frac{7}{10}[/tex] - [tex]\frac{17}{20}[/tex]
PFC = [tex]\frac{9}{20}[/tex]
PFC = 0.45
Multiply by 100 to get in in %
The probability that a 10-year old computer still has fully functioning MB and HD is 45%
The probability that a 10-year old computer still has fully functioning MB and HD is 0.45 or 45%.
Given
Probability of bad motherboard(MB) = 40% = 0.40
Probability of bad hard-drive(H) = 30%= 0.30
Probability of bad motherboard and hard-drive (MB and H) = 15% = 0.15.
What is formula is used to find the probability that a 10-year old computer still has fully functioning MB and HD?
The formula is used to find the probability that a 10-year old computer still has fully functioning MB and HD is;
[tex]\rm P(MB\ U\ H)= P(1- MB) + P(1- H) -P(1- MB \ and \ H)[/tex]
Substitute all the values in the formula;
[tex]\rm \text{P(MB U H)}= \text{P(MB)} + \text{P(H) }-\text{P(MB and H)}\\\\\text{P(MB U H)}=(1- 0.40) + (1-0.30)-(1-0.15)\\\\\text{P(MB U H)}= 0.60+0.70-0.85\\\\\text{P(MB U H)}= 0.130-0.85\\\\\text{P(MB U H)}= 0.45[/tex]
Hence, the probability that a 10-year old computer still has fully functioning MB and HD is 0.45 or 45%.
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