Asked by Shedrach
A person apply for a job in two firms say X&Y, the probability of his been selected in firm X is 0.7, and been rejected in firm Y is 0.5. The probability of atleast 1 of his application been rejected is 0.6. What is the probability he will be selected in one of the two firms.
Answers
Answered by
steve
P(x)=0.7,p(y)=1-0.5=0.5
p(x'y')+p(x'y)+p(xy')=0.6
p(x'y')=0.6-[(0.3*0.5) +(0.7*0.3)=0.24
p(x'y')=p(xory)'=0.24
p(x or y)=1-0.24=0.76
p(x'y')+p(x'y)+p(xy')=0.6
p(x'y')=0.6-[(0.3*0.5) +(0.7*0.3)=0.24
p(x'y')=p(xory)'=0.24
p(x or y)=1-0.24=0.76
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