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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 students has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly two. Which of the following relations are true?


Your Answer
A
p+m+c=1920
Correct Answer
B
p+m+c=2720
Correct Answer
C
pmc=110
Your Answer
D
pmc=14

Solution

The correct options are
B p+m+c=2720
C pmc=110
Let A, B and C respectively denote the events that the student passes in Maths, Physics and Chemistry.
It is given,
P(A) = m, P(B) = p and P(C) = c and
P (passing atleast in one subject)
=P(ABC)=0.75
1P(ABC)=0.75    [P(A)=1P(¯A)]and   [P(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ABC)=P(ABC)]1P(A).P(B).P(C)=0.75
A, B and C are independent events, therefore A', B' and C' are independent events.
    0.75 = 1 - (1 - m)(1 - p)(1 - c)
   0.25 = (1 - m)(1 - p)(1 - c)     . . .(i)
Also, P(passing exactly in two subjects) = 0.4
 P((AB¯C)(A¯BC)(¯ABC))=0.4 P(AB¯C)+P(A¯BC)+P(¯ABC)=0.4 P(A)P(B)P(¯C)+P(A)P(¯B)P(C)+P(¯A)P(B)P(C)=0.4
   pm(1 - c) + p(1 - m)c + (1 - p)mc = 0.4
   pm - pmc + pc - pmc + mc - pmc = 0.4   . . .(ii)
Again, P (passing atleast in two subjects) = 0.5
 P(AB¯C)+P(A¯BC)+P(¯ABC)+P(ABC)=0.5
   pm(1 - c) + pc(1 - m)+ cm(1 - p) + pcm = 0.5
    pm - pcm + pc - pcm + cm - pcm + pcm = 0.5
      (pm + pc + mc)-2pcm = 0.5  . . .(iii)
From Eq. (ii),
pm + pc + mc - 3pcm = 0.4     . . . (iv)
On solving Eqs. (iii), (iv) and (v), we get
p + m + c = 1.35 =2720
Therefore, option (b) is correct.
Also, from Eqs. (ii) and (iii), we get pmc =110
Hence, option c is correct.

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