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. $p+m+c=\frac{19}{20}$
• B. $p+m+c=\frac{27}{20}$
• C. $pmc=\frac{1}{10}$
• D. $pmc=\frac{1}{4}$

Answer 1

Answer: (B), (C)

The correct options are
B $p+m+c=\frac{27}{20}$
C $pmc=\frac{1}{10}$

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)
$=P(A\cup B\cup C)=0.75$
$\Rightarrow 1-P(A^{\prime }\cap B^{\prime }\cap C^{\prime })=0.75$
$\because \left[P\right(A)=1-P^{\prime }(A\left)\right]$
$\Rightarrow 1-P\left(A^{\prime }\right).P\left(B^{\prime }\right).P\left(C^{\prime }\right)=0.75$
$\because$ A, B and C are independent events, therefore A', B' and C' are independent events.
$\Rightarrow$ $0.75=1-(1-m)(1-p)(1-c)$
$\Rightarrow$ $0.25=(1-m)(1-p)(1-c)$
$\Rightarrow 0.25=1-p-m+mp-c+cp+cm-mpc$
$\Rightarrow 0.25=1-(m+p+c)+(mp+pc+cm)-mpc$ .. . . .(i)
Also, P(passing exactly in two subjects) $=0.4$
$\Rightarrow P\left(\right(A\cap B\cap \overline{C})\cup (A\cap \overline{B}\cap C)\cup (\overline{A}\cap B\cap C\left)\right)=0.4$
$\Rightarrow P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)=0.4$
$\Rightarrow P\left(A\right)P\left(B\right)P\left(\overline{C}\right)+P\left(A\right)P\left(\overline{B}\right)P\left(C\right)+P\left(\overline{A}\right)P\left(B\right)P\left(C\right)=0.4$
$\Rightarrow$ $pm(1-c)+p(1-m)c+(1-p)mc=0.4$
$\Rightarrow$ $pm-pmc+pc-pmc+mc-pmc=0.4$ . . .(ii)
Again, P (passing atleast in two subjects) $=0.5$ [Since $P\left(\overline{A}\right)=P^{\prime }\left(A\right)$]
$\Rightarrow P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)+P(A\cap B\cap C)=0.5$
$\Rightarrow P\left(A\right).P\left(B\right).P^{\prime }\left(A\right)+P\left(A\right).P^{\prime }\left(B\right).P\left(C\right)+P\left(A\right).P\left(B\right).P\left(C\right)=0.5$
$\Rightarrow$ $pm(1-c)+pc(1-m)+cm(1-p)+pcm=0.5$
$\Rightarrow$ $pm-pcm+pc-pcm+cm-pcm+pcm=0.5$
$\Rightarrow$ $(pm+pc+mc)-2pcm=0.5$ . . .(iii)
From Eq. (ii),
$pm+pc+mc-3pcm=0.4$ . . . (iv)
On solving Eqs. (iii), (iv):
From equation(iii)$-$ equation(iv), we get
$\Rightarrow (pm+pc+mc)-2pcm-pm-pc-mc+3pcm=0.5-0.4$
$\therefore pcm=0.1=\frac{1}{10}$........(v)
Thus, option (c) is correct.
Substitute $pcm=0.1$ in equation(iv), we get
$pm+pc+mc-3pcm=0.4$
$\Rightarrow pm+pc+mc-3\left(0.1\right)=0.4$
$\Rightarrow pm+pc+mc=0.4+0.3$
$\therefore pm+pc+mc=0.7$.........(vi)
Substitute equation(v) and (vi) in equation (i), we get
$\Rightarrow 0.25=1-(m+p+c)+0.7-0.1$
$\Rightarrow m+p+c=-0.25+1.6$
$\therefore m+p+c=1.35=\frac{27}{20}$
Thus, the correct option is (B) ($\frac{27}{20}$)