ProbabilityHard
Question
The probabilities that a student passes in Mathematics, phtsis and Chemistry are m.p and c respectively. Of these subjects, the students has a 75% chance of passing in at least one, a 50 % chance of of passing in at least two, and a 40% chance of passing in exactly two. Which of the following relations are true ?
Options
A.p + m + c = 
B.p + m + c = 
C.pmc = 
D.pmc = 
Solution
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 in at least one′s) = P(A ∪ B ∪ C) = 0.75
⇒ P(A′ ∩ B′ ∩ C′) = 0.75
∵ [P(A) = 1 - P
and 
= P(A ∩ B ∩ C)]
⇒ 1 - P(A′).P(B′).P(C′) = 0.75
∵ A,B,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 ∩
∪ A ∩ 
∩ C ∪ 
∩ B ∩ C) = 0.4
⇒ P(A ∩ B ∩) + P(A ∩ ∩ C) + P( ∩ B ∩ C) = 0.4
⇒ P(A)P(B)P() + P(A)P()P(C) + P(A)P(B)P() = 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 at least in two subjects) = 0.5
⇒ P(A ∩ B ∩) + P(A ∩ ∩ C) + P(
It is given :
P(A) = m, P(B) = p and P(C) = c and P (passing in at least one′s) = P(A ∪ B ∪ C) = 0.75
⇒ P(A′ ∩ B′ ∩ C′) = 0.75
∵ [P(A) = 1 - P
and = P(A ∩ B ∩ C)]⇒ 1 - P(A′).P(B′).P(C′) = 0.75
∵ A,B,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 ∩
∪ A ∩ ∩ C ∪ ∩ B ∩ C) = 0.4⇒ P(A ∩ B ∩
) + P(A ∩ ∩ C) + P( ∩ B ∩ C) = 0.4⇒ P(A)P(B)P(
) + P(A)P()P(C) + P(A)P(B)P() = 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 at least in two subjects) = 0.5
⇒ P(A ∩ B ∩
) + P(A ∩ ∩ C) + P(Create a free account to view solution
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