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

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 $6$