Question

6 (5 pts) There are 5 defective parts and 7 good parts in a box. When we elect three o from these 12 parts at random, what is the probability that we select one defective wo good parts? We do NOT replace the selected parts to the box. part and 7 (5 ptoj 3rd trial? Show all caloulations t rotl a die tunmi t gost the frst ais. What is the probability thar I eet the fest wiv at any e teacher's salary in a part (5 pts) The averag is $10,200, find the salary 8 icular state is 354.166. If the standard tertation ofzere, i,c., ?-o, show calculations or explain your solutionoepondin t (5 pts) The probability that a given tourist goes to the probability that she goes to the water part is o.St. Irtenusement partk is 0.51. If the probability that she visits both of the parks is 0.1, what is the probability thar she goes to eith s is 0.1, w 9 the water park or the amusement

Answer 1

Question 6:

There are 5 defective parts and 7 good parts.

Probability that we select one defective part and two good parts when 3 parts are selected from 12 parts is computed as:

= Number of ways to select 1 defective part from 5 defective parts * Number of ways to select 3 good parts from 7 good parts / Total number of ways to select 3 parts from 12 parts

$= \frac{\binom{5}{1}\binom{7}{2}}{\binom{12}{3}} = \frac{5*21}{220} = 0.4773$

Therefore 0.4773 is the required probability here.