P(B∪C∪F)=1−P(Bc∩Cc∩Fc)=1−0.06=0.94
P(B∪C∪F)=P(B)+P(C)+P(F)−P(B∩C)−P(B∩F)−P(F∩C)+P(B∩C∩F)
0.94=0.6+0.52+0.45−0.3−0.28−0.25+P(B∩C∩F)
P(B∩C∩F)=0.2
The number of students enrolled in 3 subjects:
1000(0.2)=200 The number of students who enrolled in only one subject:
1000((0.6−0.3−0.28+0.2)+(0.52−0.3−0.25+0.2)+(0.45−0.25−0.28+0.2))=510
The number of students who enrolled in at least 2 of 3 subjects:
940−510=430
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