If a sample of 6 keyboards is randomly selected, what is the probability that at least 5 of these will have a mechanical defect

0.1978 is the probability that at least 5 keyboards will have mechanical defect.

Step-by-step explanation:

Solution:

Given:

A repair facility has 25 failed keyboards.

10 keyboards have electrical defect.

15 of which have mechanical defect.

Randomly selected keyboard= 6

The set of A is the set of keyboards with electrical defect.

And  the set of B is set of keyboards with mechanical defect.

N(A U B) = n (A) + n(B) – n (A ∩B)

Probability of at least 5 will have mechanical defect is:

                              = p(  5 mechanical defect  U 6 mechanical defect)

                              = p(5 mechanical defect  ) + p(6 mechanical defect)

                                =(15 C 5 x 10 C1 + 15 C6 X 10C0) / 25 C6

                                = 3003 X 10 / 177100 + 5005 X 1 / 177100

                                = 30030 + 5005 / 177100

                               = 35035 / 177100

                              = 0.1978

0.1978 is the probability that at least 5 keyboards will have mechanical defect.

translate k to n and reflect across the line containing jk.

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497