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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, 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
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B
p+m+c=2720
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C
pmc=110
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D
pmc=14
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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.

189726_141229_ans.bmp

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