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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

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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

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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