Answer:
Step-by-step explanation:
(a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3ab² + 3cb² +3 ac² + 3bc² + 6abc
a = 5a ; b =y ; c = z
(5x + y + z)(5z + y + z )(5z + y +z) = (5x + y +z)³
= (5x)³ + y³ +z³ + 3(5x)²y + 3(5x)²z + 3(5x)*y² + 3*z*y² + 3*5x*z² + 3*y*z² + 6*5x*y*z
= 125x³ + y³ +z³ + 3*25x²y + 3*25x²*z + 15xy² + 3zy² + 15xz² + 3yz² + 35xyz
= 125x³ + y³ + z³ + 75x²y + 75x²z + 15xy² + 3zy² + 15xz² + 3yz² + 35xyz
Answer:
P=0.147
Step-by-step explanation:
As we know 80% of the trucks have good brakes. That means that probability the 1 randomly selected truck has good brakes is P(good brakes)=0.8 . So the probability that 1 randomly selected truck has bad brakes Q(bad brakes)=1-0.8-0.2
We have to find the probability, that at least 9 trucks from 16 have good brakes, however fewer than 12 trucks from 16 have good brakes. That actually means the the number of trucks with good brakes has to be 9, 10 or 11 trucks from 16.
We have to find the probability of each event (9, 10 or 11 trucks from 16 will pass the inspection) . To find the required probability 3 mentioned probabilitie have to be summarized.
So P(9/16 )= C16 9 * P(good brakes)^9*Q(bad brakes)^7
P(9/16 )= 16!/9!/7!*0.8^9*0.2^7= 11*13*5*16*0.8^9*0.2^7=approx 0.02
P(10/16)=16!/10!/6!*0.8^10*0.2^6=11*13*7*0.8^10*0.2^6=approx 0.007
P(11/16)=16!/11!/5!*0.8^11*0.2^5=13*21*16*0.8^11*0.2^5=approx 0.12
P(9≤x<12)=P(9/16)+P(10/16)+P(11/16)=0.02+0.007+0.12=0.147
