about to only mathematics

Options

A. p+m+c`=(19)/(20)`
B. p+m+c`=(27)/(20)`
C. `pmc=(1)/(2)`
D. `pmc=(1)/(2)`

Answer

B, C

Explanation

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 P(C) = c and
P (passing atleast in one subject)
`= P(A uu B uu C) = 0.75`
`rArr 1-P(A' nn B' nn C') = 0.75`
`because [P(A) = 1-P(bar(A))`
and `[P(bar(A uu B uu C)] = P(A' nn B' nn C')]`
`rArr 1-P(A') * P(B') * P(C') = 0.75`
`because ` A, B and C are independent events, therefore A', B' and C' are independent events.
`rArr 0.75 = 1-(1-m)(1-p) (1-c)`
`rArr 0.25 = (1-m)(1-p) (1-c) " " ...(i)`
Also, P (passing exactly in two subjects) = 0.4
`rArr P(A nn B nn bar(C) nn A nn bar(B) nn C uu bar(A) nn B nn C) = 0.4`
`rArr P(A nn B nn bar(C)) + P(A nn bar(B) nn C) + P(bar(A) nn B nn C) = 0.4`
`rArr P(A) P(B) P(bar(C)) + P(A)P(bar(B)) P(C) + P (bar(A)) P(B) P(C) = 0.4`
`rArr pm (1-c) + p(1-m)c + (1-p)mc = 0.4 `
`rArr pm - pmc + pc- pmc + mc -pmc = 0.4 " "...(i)`
Again P(passing alteast in two subjects) = 0.5
`rArr P(A nn B nn bar(C)) + P(A nn bar (B) nn C) + P(bar(A) nn B nn C) + P(A nn B nn C) = 0.5`
`rArr pm (1-c) + pc(1-m) +cm(1-p) + pcm = 0.5`
`rArr pm-pcm + pc -pcm + cm -pcm +pcm = 0.5`
`rArr (pm +pc + mc) - 2pcm = 0.5 ...(iii)`
From Eq. (ii),
`pm + pc +mc -3pcm = 0.4 .... (iv)`
From Eq. (i),
`0.25 = 1-(m +p+c) + (pm +pc + cm) -pcm ....(iv)`
On solving Eqs. (iii), (iv), and (v), we get
p +m +c = 1.35 = 24/20
Therefore, option (b) is correct.
Also, from Eqs. (ii) and (iii), we get pmc = 1/10
Hence, option (c) is correct.