Question

In an examination, one hundred candidates took papers in Physics and Chemistry. Twenty five candidates failed in Physics only. Twenty candidates failed in Chemistry only. Fifteen failed in both Physics and Chemistry. A candidate is selected at random. The probability that he failed either in Physics or in Chemistry but not in both, is
A) $\dfrac{9}{{20}}$
B) $\dfrac{3}{5}$
C) $\dfrac{2}{5}$
D) $\dfrac{{11}}{{20}}$

Answer 1

Here we need the probability that the candidate failed either in Physics or in Chemistry but not in both. For that, we will use the addition property on probability. For that, we will substitute all the data given in the question in the formula. From there, we will get the value of required probability.

Complete step by step solution:
Here we need the probability that the candidate failed either in Physics or in Chemistry but not in both.
We know from addition property of probability that if $A$ and $B$ are two events which is associated with a random experiment, then $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$.
It is given that:
Number of students failed in both chemistry and physics $= 15$
Number of students failed in Physics only $= 25$
But the total number of students who failed in physics is equal to the sum of the number of students who failed in Physics only and the number of students failed in both chemistry and physics.
Therefore, total number of students failed in physics $= 25 + 15 = 40$
Number of students failed in chemistry only $= 20$
But the total number of students who failed in chemistry is equal to the sum of the number of students who failed in chemistry only and the number of students failed in both chemistry and physics.
Therefore, total number of students failed in chemistry $= 20 + 15 = 35$
Total number of candidates $= 100$
Now, we will find the probability that the students failed in Physics, probability of students failed in chemistry and the probability that the students failed in both chemistry and physics.
Let the probability that the students failed in Physics be $P\left( P \right)$, $P\left( C \right)$ be probability of students failed in chemistry and therefore, the probability that the students failed in both chemistry and physics will be $P\left( {P \cap C} \right)$
Therefore,
$\Rightarrow P\left( P \right) = \dfrac{{40}}{{100}}$
$\Rightarrow P\left( C \right) = \dfrac{{35}}{{100}}$
$\Rightarrow P\left( {P \cap C} \right) = \dfrac{{15}}{{100}}$
Now, we will calculate the probability that the students failed either in Physics or in Chemistry but not in both, it will be represented by $P\left( {P \cup C} \right)$.
Using the addition property of probability, we get
$P\left( {P \cup C} \right) = P\left( P \right) + P\left( C \right) - P\left( {P \cap C} \right)$
Now, we will substitute the value of all probabilities obtained in this formula. Therefore, we get
$\Rightarrow P\left( {P \cup C} \right) = \dfrac{{40}}{{100}} + \dfrac{{35}}{{100}} - \dfrac{{15}}{{100}}$
On adding and subtracting the numbers, we get
$\Rightarrow P\left( {P \cup C} \right) = \dfrac{{60}}{{100}}$
On further simplification, we get
$\Rightarrow P\left( {P \cup C} \right) = \dfrac{3}{5}$
The probability that the candidates failed either in Physics or in Chemistry but not in both is equal to $\dfrac{3}{5}$

Hence, the correct option is option B.

Note:
To solve this question, we need to have knowledge about probability and its properties. Probability is defined as the ratio of number of desired or favorable outcomes to the total number of possible outcomes. Also, the value of probability cannot be greater than 1 and the value of probability cannot be negative. In addition to this, the probability of a sure event is always one.