Suppose that after 10 years of service, 40% of computers have problems with motherboards (mb), 30% have problems with hard drives (hd), and 15% have problems with both mb and hd. what is the probability that a 10-year old computer still has fully functioning mb and hd?

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

PFMB = 1 - probability of having problems with motherboards = 1 - 40 %

PFMB  = 1 - = 1 - =

PFHD = 1 - probability of having problems with hard drive = 1 - 30 %

PFHD = 1 - = 1 - =

PFMH = 1 - probability of having problems with both = 1 - 15 %

PFMH = 1 - = 1 - =

PFC = + -

PFC =

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%

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0

30,40,50 is correct answer