Question

The probability that a student passes in mathematics, physics and chemistry are m, p and c respectively. Of these subjects, a student 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 subjects. 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. \[pmc = \dfrac{1}{4}\]

Answer 1

Hint: In this question probability for three conditions are given so first we will write these conditions in the reduced form and then by solving those equations we will find the best relation between \[p,c,m\]
The probability of passing in at least two subjects can also be written as probability of passing in two subjects plus probability of passing in all three subjects.

Complete step-by-step answer:
Given
The probability of student passing in mathematics is \[m\]
In physics is \[p\]
In chemistry is \[c\]
Also the chance of passing in at least one subject \[ = 75\% \]
The chance of passing in at least two subjects \[ = 50\% \]
The chance of passing in exactly two subjects \[ = 40\% \]
Now we can say the probability of not passing in any one subject can be written as \[ = 1 - P\] (passing in at least one subject), where the probability of passing in one subject is given as \[ = 75\% \] , hence we can write
 \[P\] (Not passing in one subject) \[ = 1 - \dfrac{{75}}{{100}} = \dfrac{{25}}{{100}}\]
This can also be written as
 \[\left( {1 - p} \right)\left( {1 - m} \right)\left( {1 - c} \right) = \dfrac{{25}}{{100}}\]
By solving this
 \[
  \left( {1 - p - m + pm} \right)\left( {1 - c} \right) = \dfrac{{25}}{{100}} \\
   \Rightarrow 1 - p - m + pm - c + pc + mc - pmc = 0.25 \\
   \Rightarrow p + m + c - pm - pc - mc + pmc = 0.75 - - (i) \;
 \]
Now we are given the probability of passing in two subjects as 40% this means a student will pass in any two of the given three subjects and will fail in the third one, hence we can also write this as
 \[
  mp\left( {1 - c} \right) + mc\left( {1 - p} \right) + pc\left( {1 - m} \right) = \dfrac{{40}}{{100}} \\
   \Rightarrow mp - mpc + mc - mpc + pc - mpc = 0.4 \\
   \Rightarrow mp + pc + cm - 3mpc = 0.4 - - (ii) \;
 \]
Now we are given the probability of passing in at least two subjects is 50%, this can be written in the form of (probability of passing in two subjects + probability of passing in three subjects), hence we can write this in reduce form from equation (ii) as
 \[
  mp + pc + cm - 3mpc + mpc = \dfrac{{50}}{{100}} \\
   \Rightarrow mp + pc + cm - 2mpc = 0.5 - - (iii) \;
 \]
Now subtract (ii) from (iii), we get
 \[
  mp + pc + cm - 2mpc - mp - pc - cm + 3mpc = 0.5 - 0.4 \\
   \Rightarrow mpc = 0.1 - - (iv) \;
 \]
Now we add equation (iii) with (iv), we get
 \[
  mp + pc + cm - 2mpc + mpc = 0.5 + 0.1 \\
   \Rightarrow mp + pc + cm - mpc = 0.6 - - (v) \;
 \]
Now we add equation (v) with equation (i), we get
 \[
  p + m + c - pm - pc - mc + pmc + mp + pc + cm - mpc = 0.6 + 0.75 \\
   \Rightarrow p + m + c = 1.35 \;
 \]
Hence we can write this as
 \[
  p + m + c = \dfrac{{135}}{{100}} \\
   = \dfrac{{27}}{{20}} \;
 \]
So, the correct answer is “Option C”.

Note: If \[P\left( S \right) = p\] is the probability of success then the probability of failure \[P\left( F \right) = q\] can be written as \[q = 1 - p\].
In a binomial experiment there can only be two outcomes either the experiment is a success or the experiment is a failure. Where p denotes the probability of successful trials and q denotes the probability of unsuccessful trials.