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 $-\left(a\right)$ only one subject $\left(b\right)$ neither physics nor chemistry

Answer 1

Total students $=100$

$N\left(M\right)=45$ N(M) represents number of students studying maths

$N\left(P\right)=48$

$N\left(C\right)=40$

$N\left(MandP\right)=12,\left(MandC\right)=15,N\left(PandC\right)=11$

$N\left(MandPandC\right)=5$

student studying only maths

$=N\left(M\right)-N\left(MandP\right)-N\left(MandC\right)+N\left(M\&P\&C\right)$

$=45-12-15+5$

$=23$

student studying only physics

$=N\left(P\right)-N\left(PandM\right)-N\left(PandC\right)+N\left(P\&M\&C\right)$

$=48-12-11+5=30$

students studying only chem

$=N\left(C\right)-N\left(CandP\right)-N\left(CandM\right)+N\left(PandC\&M\right)$

$=40-11-15+5=19$

a) probability that student studies only one subject

$=\frac{19+30+23}{100}=0.72$

b) probability that student studies neither physics nor chemistry

$=$ probability that student studies only Maths $=\frac{23}{100}=0.23$