Aconsumer organization estimates that over a 1-year period 20% of cars will need to be repaired once, 5% will need repairs twice, and 1% will require three or more repairs. if you own two cars, what is the probability that

a) neither will need repair?
b) both will need repair?
c) at least one car will need repair?

a) 0.5476

b) 0.0676

c) 0.4524

Step-by-step explanation:

Given this information, we can conclude that 74% of the cars won't need any repairs over a 1-year period (100 - 20 - 5 - 1 = 74%). And 26% will need at least 1 repair over a 1-year period.

P(car doesn't need repair) = 0.74

P (car needs repair) = 0.26

If you own two cars, the probability that:

a) Neither will need repair:

We need that car 1 won't need repair AND car 2 won't need repair.

=P(Car 1 doesn't need repair) x P(Car 2 doesn't need repair)

= 0.74 x 0.74 = 0.5476

The probability that neither will need repair is 0.5476.

b) Both will need repair:

We need that car 1 needs repair AND car 2 needs repair.

P(Car 1 needs repair) x P(Car 2 needs repair)

= 0.26 x 0.26 = 0.0676

The probability that both will need repair is 0.0676

c) At least one car will need repair

Car 1 needs repair or Car 2 needs repair or both need repair.

To solve this one, it's easier to use the complement of P(neither needs repair)

1 - P(neither needs repair)

1 - (0.74)(0.74)  = 1 - 0.5476 = 0.4524

The probability that at least one car will need repair is 0.4524

b

step-by-step explanation:

given

- 2 (3x + 12y - 5 - 17x - 16y + 4 )

simplify the parenthesis by collecting like terms

= - 2 (- 14x - 4y - 1) ← distribute the parenthesis by - 2

= 28x + 8y + 2 → b