4percent of all cars produced on a monday are deficient, while 3 percent of friday's production is deficient; on all other days, only 1 percent is defective. let a car chosen randomly from the production be deficient. what is the probability that it was produced on monday? what is the solution if 12 percent of the cars are produced on friday and the remaining production is distributed equally over the other 4 working days?

Let the total numbers of cars be ten then the total cars produced in one week would be 50

Probability of deficient cars on Monday = 4/10

Probability of deficient cars on Tuesday =1/10

Probability of deficient cars on Wednesday = 1/10

Probability of deficient cars on Thursday = 1/10

Probability of deficient cars on Friday = 3/10

Probability ( deficient car)= 10/50 or 1/5

Probability that it produced on Monday 4/10*1/5= 4/50= 2/25= 8/100= 0.08 or 8%

Probability of deficient cars on Monday = 4/10* 12/100= 48/1000= 0.048or 4.8%

Probability of deficient cars on Tuesday =1/10 *22/100= 22/1000= 0.022 or 2.2%

Probability of deficient cars on Wednesday = 1/10*22/100=22/1000= 0.022 or 2.2%

Probability of deficient cars on Thursday = 1/10*22/100=22/1000= 0.022 or 2.2%

Probability of deficient cars on Friday = 3/10*22/100= 66/1000= 0.066 or 6.6%

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.