1

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?

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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(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.

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(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.

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