Question

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. The probability that the student fails in the examination is
$A)\dfrac{{197}}{{200}}$
$B)\dfrac{{27}}{{100}}$
$C)\dfrac{{83}}{{100}}$
$D)$ None of these

Answer 1

Hint:
> 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)
> Now we have three known values which are division I, II and III passed students.
> Add the all division passed students probability and subtracted from the number $1$ will get the resultant.
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, probabilities of the student getting I, II, and III division in an examination are $\dfrac{1}{{10}},\dfrac{3}{5},\dfrac{1}{4}$ respectively.
Which can be represented as 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 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 \[P(F) = (1 - \dfrac{1}{{10}}) \times (1 - \dfrac{3}{5}) \times (1 - \dfrac{1}{4}) = \dfrac{9}{{10}} \times \dfrac{2}{5} \times \dfrac{3}{4} =\dfrac{{54}}{{200}} = \dfrac{{27}}{{100}}\] (canceling the common terms)
Hence option $B)\dfrac{{27}}{{100}}$ is correct.

Note:
> Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of the outcomes.
> As we mention in the hint, the overall total amount that students passed and failed is $1$ (by the probability rule).
> 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 $6$