Question

In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first exam is $0.8$ and the probability of passing the second exam is $0.7$. The probability of passing at least one of the exams is $0.95$.What is the probability of passing both?

Answer 1

Probability of any event is represented as $P\left( {event} \right)$. If $A$ and $B$ be the two events then $P\left( {A \cup B} \right)$ represents the probability of getting either $A$ or $B$ at a time and $P(A \cap B)$ represents the probability of getting both $A$ and$B$. Addition rule of probability establishes the relation between them as $P\left( {A \cap B} \right) = P\left( A \right){\text{ }} + {\text{ }}P\left( B \right){\text{ }}-{\text{ }}P\left( {A \cup B} \right)$.

Complete step by step answer:
We have given the probability of a random chosen student passing the first exam, second exam and passing at least one exam. We have to find the probability of passing both the exams. Let $A$ be the event of passing the first exam by a random chosen student, $B$ be the event of passing the second exam. Then,
Probability of passing the first exam is given by, $P\left( A \right) = 0.8$
Probability of passing the first exam is given by, $P\left( B \right) = 0.7$

Probability of passing at least one exam, that is passing either of exams is given by $P\left( {A \cup B} \right) = 0.95$
Probability of passing both the exams is given by, $P\left( {A \cap B} \right) = ?$
Now from addition rule of probability we have,
$P\left( {A \cap B} \right) = P\left( A \right){\text{ }} + {\text{ }}P\left( B \right){\text{ }}-{\text{ }}P\left( {A \cup B} \right)$
Substituting the given values we have,
$P\left( {A \cap B} \right) = 0.8 + 0.7 - 0.95$
On simplifying we get
$P\left( {A \cap B} \right) = 1.5 - 0.95$
On subtracting we get,
$\therefore P\left( {A \cap B} \right) = 0.55$

Hence, Probability of passing both exams by a randomly chosen student is $0.55$.

Note: Probability of any event cannot be less than $0$ or greater than $1$, it lies in between $0$ and $1$. Probability has no unit as it is a ratio. Probability of getting $A$ and $B$ together for two independent variables $A$ and $B$ is given as $P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$. In probability ‘or’ is denoted as Union and ‘and’ is denoted as intersection. The addition rule is derived from set theory.