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Question

# The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 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. Following relations are drawn in m, p, c I. p+m+c=2720 II. p+m+c=1320 III. (p)×(m)×(c)=110

A
Only relation I is true.
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B
Only relation III is true.
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C
Relations II and III are true.
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D
Relations I and III are true.
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Solution

## The correct option is D Relations I and III are true. Given, α+β+γ+d+e+f+g=75100 ...(1) & d+e+f+g=50100 ...(2) & d+e+f=40100 ...(3) Subtracting (2) & (3) g=10100 = 10% & subtracting (1) & (2), α+β+γ=25100 So M × P × Ch = 10% Now, M + P + Ch ⇒ (α+d+g+f)+(β+d+e+g)+(γ+e+f+g) =(α+β+γ)+2(d+e+f)+3g =25100+2(40100)+3(10100) =135100=2720 Method - II % of students passing in atleast one subject = 75% i.e. p+m+c−{p∩m+m∩c+c∩p}+p∩m∩c=0.75 ...(1) % of students passing in at least two subjects = 50% {p∩m+m∩c+c∩p}−2{p∩m∩c}=0.5 ...(2) % of students passing in exactly two subjects = 40% {p∩c+m∩c+c∩p}−3{p∩m∩c}=0.4 ...(3) By (2) and (3) p∩m∩c=0.1⇒p×m×c=110 Adding (1) and (2) p+m+c−(p∩m∩c)=1.25 p+m+c=1.25+0.1=1.35=135100=2720 Hence option (d) is correct

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