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Question

In a class, there are 100 students out of which 45 study mathematics, 48 study physics, 40 study chemistry, 12 study both mathematics and physics, 11 study both physics and chemistry, 15 study both mathematics and chemistry and 5 study all three subjects. A student is selected at random, then find the probability that the selected student studies −(a) only one subject (b) neither physics nor chemistry

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Solution

Total students =100

N(M)=45 N(M) represents number of students studying maths

N(P)=48

N(C)=40

N(M and P)=12,(M and C)=15,N(P and C)=11

N(M and P and C)=5

student studying only maths

=N(M)−N(M and P)−N(M and C)+N(M & P & C)

=45−12−15+5

=23

student studying only physics

=N(P)−N(P and M)−N(P and C)+N(P & M & C)

=48−12−11+5=30

students studying only chem

=N(C)−N(C and P)−N(C and M)+N(P and C & M)

=40−11−15+5=19

a) probability that student studies only one subject

=19+30+23100=0.72

b) probability that student studies neither physics nor chemistry

= probability that student studies only Maths =23100=0.23

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