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 at least one, a 50% chance of passing in at least 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

The Correct Option is C

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)
$\hspace30mm =P(A \cup B \cup C)=0.75 $
$\Rightarrow \, \, \, \, 1-P(A' \cap B' \cap C')=0.75$
$\because \, \, \, \, \, \, \, \, \, \, \, \, \, \, [P(A)=1-P(\overline{A}) $
and $ \, \, \, \, \, \, \, \, [P\overline{(A \cup B \cup C)}]=P(A' \cap B' \cap C')]$
$\Rightarrow \, \, \, \, \, \, \, \, \, 1-P(A').P(B').P(C')=0.75$
$\therefore \, A, B$ and $C$ are independent events, therefore $A', B' $ and $C $' are independent events.
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 0.75=1-(1-m)(1-q)(1-c)$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 0.25=(1-m)(1-q)(1-c) $
...(i)
Also, P (passing exactly in two subj|ects)= 0.4
$\Rightarrow \, P(A \cap B \cap \overline{C} \cup \, A \cap \overline{B} \cap C \cup \overline{A} \cap B \cap C)= 0.4$
$\Rightarrow \, \, P(A \cap B \cap \overline{C}) +\, P(A \cap \overline{B} \cap C ) + \, P(\overline{A} \cap B \cap C)=0.4$
$\Rightarrow \, \, P(A)P(B)P(\overline{C})+P(A)P(\overline{B})P(C)$
$\hspace30mm + P(\overline{A}) P(B) P(C)= 0.4$
$\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
$\hspace30mm + P(\overline{A}) P(B) P(C)= 0.4$
$\Rightarrow \, \, \, pm(1 - c) + pc(1 -m )+ cm( 1 - p) + pcm = 0.5$
$\Rightarrow \, \, \, pm - pcmn + p c - pcm + cm - pcm + pcm = 0.5 $
$\Rightarrow \, \, \, (pm + pc + me ) -2pcm =0.5 ...........$(iii)
From E (ii),
$\hspace20mm pm+ pc+ mc-3pcm =0.4 ...$(iv)
From E (i),
$025 = 1 - (m + p + c )+ (pm + pc + cm) - pcm ........... $.(v)
On solving Eqs. (iii), (iv) and (v), we get
$\hspace20mm p + m + c= 1.35 = 27/20$
Therefore, option (b) is correct.
Also, from Eqs. (ii) and (iii), we get pmc= 1/10

Hence, option (c) is correct.