  Question

The probabilities that a student passes in mathematics, physics and chemistry are $$m, p$$ and $$c$$ respectively. Of these subjects, a student has a $$75%$$ chance of passing in at least one, a $$50%$$ chance of passing in at least two, and a $$40%$$ chance of passing in exactly two subjects. Which of the following relations are true?

A
p+m+c=1920  B
p+m+c=2720  C
pmc=110  D
pmc=14  Solution

The correct options are B $$\displaystyle p + m + c= \frac {27}{20}$$ C $$\displaystyle pmc= \frac {1}{10}$$Referring to the illustration:Let s, t, u, v, w, x, y and z are the probabilities of the respective regions.Thus, we have:$$s+t+u+v+w+x+y=0.75$$ ...(i)$$v+w+x+y=0.5$$ ...(ii)$$v+w+x=0.4$$ ...(iii)Thus, subtracting (ii) and (iii): $$y=pmc=0.1=\frac{1}{10}$$Here, y refers to the intersection of all the three and hence equals $$p*c*m$$ since they are independent events.Again, subtracting (i) and (ii): $$s+t+u=0.25$$We need to find $$m+c+p$$:$$m+c+p=s+t+u+2(v+w+x)+3y= 0.25+2*0.4+3\times 0.1=1.35=\dfrac { 27 }{ 20 }$$Hence, (b), (c) are correct. Maths

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