  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?

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

Solution

## The correct options are B p+m+c=2720 C pmc=110Let 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(A∪B∪C)=0.75 ⇒1−P(A′∩B′∩C′)=0.75∵    [P(A)=1−P(¯A)]and   [P(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯A∪B∪C)=P(A′∩B′∩C′)]⇒1−P(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((A∩B∩¯C)∪(A∩¯B∩C)∪(¯A∩B∩C))=0.4⇒ P(A∩B∩¯C)+P(A∩¯B∩C)+P(¯A∩B∩C)=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(A∩B∩¯C)+P(A∩¯B∩C)+P(¯A∩B∩C)+P(A∩B∩C)=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.  Suggest corrections   