Question

If the probability of the occurrence of an event is $1 - P$ then what is the probability that the event doesn’t occur.
$1)P$
$2)P + 1$
$3)1 - P$
$4)$ None of these

Answer 1

Hint: Probability is the term with events that occur, which is the number of favorable events that divides the total number of the outcomes.
If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
$\dfrac{1}{6}$ which means the favorable event is $1$ and the total outcome is $6$
Formula used:
$P = \dfrac{F}{T}$where P is the overall probability, F is the possible favorable events and T is the total outcomes from the given.

Complete step-by-step solution:
Since from the given that we have the probability of the occurrence of a given event is $1 - P$.
And we need to find the probability that the event doesn’t occur.
First, let us assume the overall total probability value is $1$ (this is the most popular concept that used in the probability that the total fraction will not exceed $1$and everything will be calculated under the number $0 - 1$ as zero is the least possible outcome and one is the highest outcome)
Therefore, the probability of the given event total is $1$
Thus, the probability of the event $1 - P$ that doesn’t occur is $1 - (1 - P)$ hence we have $1 - 1 + P = P$
Therefore, the option $1)P$ is correct.

Note: Example: The probabilities of the student getting I, II, and III division in an examination are $\dfrac{1}{{10}},\dfrac{3}{5},\dfrac{1}{4}$ respectively. Then the probability that the student fails in the examination is
Which can be represented as the probability of student getting division, I is $P(I) = \dfrac{1}{{10}}$, probability of student getting division II is $P(II) = \dfrac{3}{5}$, and probability of student getting division III is $P(III) = \dfrac{1}{4}$
Let us add all the probability we get, $P(I) \times P(II) \times P(III) = \dfrac{1}{{10}} \times \dfrac{3}{5} \times \dfrac{1}{4}$ (all possible)
Since the requirement is the failure of students which means they did not pass the exam, so subtract each and every probability with the $1$ (overall probability) we get $P(F) = (1 - \dfrac{1}{{10}}) \times (1 - \dfrac{3}{5}) \times (1 - \dfrac{1}{4})$
Further solving we get \[P(F) = (1 - \dfrac{1}{{10}}) \times (1 - \dfrac{3}{5}) \times (1 - \dfrac{1}{4}) \Rightarrow \dfrac{9}{{10}} \times \dfrac{2}{5} \times \dfrac{3}{4} \Rightarrow \dfrac{{54}}{{200}} = \dfrac{{27}}{{100}}\] (canceling the common terms)