# Difficult Probability Solved QuestionAptitude Discussion

 Q. The probabilities that a student pass in Mathematics, Physics and Chemistry are $m$, $p$, $c$ respectively. Of these subjects the student has a 75 % chance of passing in at least one, a 50% chance of passing in at least two and 40 % chance of passing in exactly two.   Which of the following relations are true ?
 ✖ A. $p+m+c=\dfrac{19}{20}$ ✔ B. $p+m+c=\dfrac{27}{20}$ ✖ C. $pmc=\dfrac{1}{10}$ ✖ D. $pmc=\dfrac{1}{4}$

Solution:
Option(B) is correct

Consider the figure shown below.

$p \cup m \cup c = 0.75$

$w + x + y = 0.4$

$z = 0.1$

$p \cup m \cup c = p + m + c - (w + x + y + 3z) + z$

$p + m + c = 1.35 =\dfrac{27}{20}$

Edit: Thank you Arjun for alternate explanation in the comments.

## (4) Comment(s)

Ashok
()

Roop
()

Arjun's explanation is lucidly presented than the answer...😊

Arjun
()

Probability of non pass = 1 - Probability of at least one pass

$= 1 - 0.75 = 0.25$

$(1-m) (1-p) (1-c) = 0.25$

$(1 + mp - m -p) (1-c) = 0.25$

$1 + mp - m - p - c - mpc + mc + pc = 0.25$

$m + p + c - mp - pc - mc + mpc = 0.75$ -- (1)

Probability of exactly 2 pass $= 0.4$

$mp (1-c) + pc (1-m) + mc (1-p) = 0.4$

$mp + pc + mc - 3mpc = 0.4$

$mp + pc + mc - 2mpc = 0.5$ -- (2) (Adding the probability of all pass to probability of exactly 2 pass gives probability of at least 2 pass)

So, $mpc = 0.1$, -- (3)

From (2) and (3),

$mp + pc + mc - mpc = 0.6$ -- (4)

From (1) and (4),

$m + p + c = 0.75 + 06$

$m + p + c = 1.35$

$= \dfrac{135}{100}$

$= \dfrac{27}{20}$

Amal
()

Both options B & C are correct.

p.m.c = 0.1 in the answer itself