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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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