Question

# In a class of $$60$$ students, $$30$$ opted for Mathematics, $$32$$ opted for Biology and $$24$$ opted for both Mathematics and Biology. If one of these students is selected at random, find the probability that: (i) The student opted for Mathematics or Biology. (ii) The student has opted neither Mathematics nor Biology. (iii) The student has opted Mathematics but not Biology.

Solution

## Number of students opted mathematics $$=n(M)=30$$ Number of students opted Biology $$=n(B)=32$$ Number of students opted both Mathematics and Biology $$=n\left( M\cap B \right) =24$$ Number of students opted Mathematics or Biology $$=\left( M\cup B \right)$$ $$n\left( M\cup B \right) =n(M)+n(B)-n\left( M\cap B \right) \\ \Rightarrow n\left( M\cup B \right) =30+32-24=38$$ (i) Probability that student opted Mathematics or Biology $$=\dfrac { n\left( M\cup B \right) }{ n } =\dfrac { 38 }{ 60 } =\dfrac { 19 }{ 30 }$$ (ii) Probability that student has opted neither mathematics nor Biology $$=1-P(M\cup B)=1-\dfrac { 19 }{ 30 } =\dfrac { 11 }{ 30 }$$ Number os students opted only Mathematics $$=n(M)-n\left( M\cap B \right) =30-24=8$$ (iii) Probability of selecting students who opted only mathematics $$=\dfrac { 8 }{ 60 } =\dfrac { 2 }{ 15 }$$  Mathematics

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