Answer
a) $P(A)=0.3$
b) $P(B)=0.77$
c) $P(A')=0.7$
d) $P(A \cap B)=0.22$
e) $P(A \cup B)=0.85$
f) $P(A'\cap B)=0.55$
Work Step by Step
Given:
Total observations = 100
A denotes sample is from supplier one.
Thus, $n(A)=22+8=30$
B denotes that sample conforms to specifications.
Thus, $n(B)=22+25+30=77$
Thus,
a) $P(A)=30/100=0.3$
b) $P(B)=n(B)/100=77/100=0.77$
c) $P(A')=1-p(A)=1-0.3=0.7$
d) $P(A \cap B)=n(A \cap B)/100$
We can see from the data that number of events where sample is from Supplier 1 (A) and conforms to the specification = 22. Thus, $n(A \cap B)=22$
Thus,
$P(A \cap B)=n(A \cap B)/100=22/100=0.22$
e) We know that: $P(A \cup B)=P(A)+P(B)-P(A\cap B)$
Thus,
$P(A \cup B)= 0.3+0.77-0.22=0.85$
f) We want to find suppliers that are not Supplier 1, ie. Supplier Two and Three, and who conform to the specifications. Thus, from the data: $n(A'\cap B)=25+30=55$
Thus,
$P(A'\cap B)=55/100=0.55$
