in a class of 130 students 85 passed in maths, 60 passed in social studies and 10 failed in both the subjects. how many students failed in exactly one subject?
If 10 failed in both, that means 130-10=120 passed at least one subject.
Since 85 passed math, 60 passed science, there are 145 passes from 120 students. We conclude that 145-120=25 passed both.
Hence (b) 120-25=95 passed (or failed) exactly one subject.
For (a), 60 passed science, and 25 passed both, so 35 passed science but did not make math.
A more mathematical treatment would be:
P'(A∪B)=1-P(A∪B) P(A)=85/130 P(B)=60/130
P'(A∪B)=1-P(A∪B)=10/130
=>
P(A∪B)=(130-10)/130=120/130
We also know that:
P(A∪B)=P(A)+P(B)-P(A∩B)
120/130=85/130+60/130-P(A∩B)
=>
P(A∩B)=85/130+60/130-120/130=25/130
The cardinality of a set is the number of members in the set.
For example, the cardinality of A is denoted by |A|, which is equal to the number of students who passed math=85.
Hence |B|=60, |A∩B|=25, |A∪B|=120.
(a)
|A’∩B|=|B|-|A∩B|=60-25=35
(b)
|A∪B|-|A∩B|=120-25=95
in a class of 130 students 85 passed in maths, 60 passed in social studies and 10 failed in both the subjects. how many students failed in exactly one subject?
If 10 failed in both, that means 130-10=120 passed at least one subject.
Since 85 passed math, 60 passed science, there are 145 passes from 120 students. We conclude that 145-120=25 passed both.
Hence (b) 120-25=95 passed (or failed) exactly one subject.
For (a), 60 passed science, and 25 passed both, so 35 passed science but did not make math.
A more mathematical treatment would be:
P'(A∪B)=1-P(A∪B)
P(A)=85/130
P(B)=60/130
P'(A∪B)=1-P(A∪B)=10/130
=>
P(A∪B)=(130-10)/130=120/130
We also know that:
P(A∪B)=P(A)+P(B)-P(A∩B)
120/130=85/130+60/130-P(A∩B)
=>
P(A∩B)=85/130+60/130-120/130=25/130
The cardinality of a set is the number of members in the set.
For example, the cardinality of A is denoted by |A|, which is equal to the number of students who passed math=85.
Hence |B|=60, |A∩B|=25, |A∪B|=120.
(a)
|A’∩B|=|B|-|A∩B|=60-25=35
(b)
|A∪B|-|A∩B|=120-25=95