Answer:
0.214 ; 0.064 ; 0.3859 ; 0.9966 ; 0.135
Step-by-step explanation:
Given the distribution :
Poor ___ Acceptable ____ Excellent
0.25 _____ 0.6 _________ 0.15
The data scientist receives three items produced by the manufacturing process.
Find the chance that:
a. all of items are Acceptable
0.6 * 0.6 * 0.6 = 0.6^3 = 0.214
b. none of the items is Acceptable
P(not acceptable) = 1 - p(acceptable) = 1 - 0.6 = 0.4
0.4 * 0.4 * 0.4 = 0.4^3 = 0.064
c. at least one of the items is Excellent
P(atleast one) = 1 - P(none)
P(not excellent) = 1 - 0.15 = 0.85
P(none is excellent) = 0.85^3 = 0.614125
Hence,
P(atleast one is excellent) = 1 - 0.614125 = 0.3859
d. not all of the items are Excellent
P(not all) = 1 - p(All)
P(all excellent) = 0.15^3 = 0.003375
P(not all are excellent) = 1 - 0.003375 = 0.9966
e. there is one item of each class:
P(poor) * p(acceptable) * p(excellent) * (number of classes)!
0.25 * 0.6 * 0.15 * 3!
= 0.135
