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)
$=P(A\cup B\cup C)=0.75$
$\Rightarrow 1-P(A^{\prime }\cap B^{\prime }\cap C^{\prime })=0.75\because \left[P\right(A)=1-P(\overline{A}\left)\right]and\left[P\right(\overline{A\cup B\cup C})=P(A^{\prime }\cap B^{\prime }\cap C^{\prime }\left)\right]\Rightarrow 1-P\left(A^{\prime }\right).P\left(B^{\prime }\right).P\left(C^{\prime }\right)=0.75$
$\because$ 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) . . .(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
$\Rightarrow P(A\cap B\cap \overline{C})+P(A\cap \overline{B}\cap C)+P(\overline{A}\cap B\cap C)+P(A\cap B\cap C)=0.5$
$\Rightarrow$ 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) and (v), we get
p + m + c = 1.35 $=\frac{27}{20}$
Therefore, option (b) is correct.
Also, from Eqs. (ii) and (iii), we get pmc $=\frac{1}{10}$
Hence, option c is correct.