Question

The odds in favor of standing first of three students appearing in an examination are $1:2$, $2:5$ and $1:7$ respectively. The probability that either of them will stand first, is
A.$\dfrac{{125}}{{168}}$
B.$\dfrac{{75}}{{168}}$
C.$\dfrac{{32}}{{168}}$
D.$\dfrac{4}{{168}}$

Answer 1

Here, we are required to find the probability of the event that either of the given three students stand first in the examination. To find the probability of the event happening, we will take the first element of the ratio as the numerator and we will take the sum of both the elements of the ratio as the denominator to showcase it as the total probability. And since, these events are mutually exclusive, we will add all the three probabilities to find the required one.

Complete step-by-step answer:
In the given question, we are provided by ‘odds in favor’ of various events.
By ‘odds in favor’, we mean that it is a ratio of the probability that the specific event will happen to the probability that it will not happen.
Now, let the three students be P, Q and R respectively and the events of them standing first in the examination be A,B and C respectively.
Now, for student P,
Odds in favour $= 1:2$
Hence, the probability of student P to stand first in the examination will be:
$P\left( A \right) = \dfrac{1}{{1 + 2}} = \dfrac{1}{3}$
Similarly, for student Q,
Odds in favour $= 2:5$
Hence, the probability of student Q to stand first in the examination will be:
$P\left( B \right) = \dfrac{2}{{2 + 5}} = \dfrac{2}{7}$
Also, for student R,
Odds in favour $= 1:7$
Hence, the probability of student R to stand first in the examination will be:
$P\left( C \right) = \dfrac{1}{{1 + 7}} = \dfrac{1}{8}$
Now the required probability that either of them will stand first, will be the sum of all three probabilities. This is because all the three events are mutually exclusive.
Therefore,
$P\left( A \right) + P\left( B \right) + P\left( C \right) = \dfrac{1}{3} + \dfrac{2}{7} + \dfrac{1}{8}$
The LCM of 3, 7 and 8 is 168. So, taking LCM on RHS, we get
$\Rightarrow P\left( A \right) + P\left( B \right) + P\left( C \right) = \dfrac{{56 + 48 + 21}}{{168}}$
Adding the terms in the numerator, we get
$\Rightarrow P\left( A \right) + P\left( B \right) + P\left( C \right) = \dfrac{{125}}{{168}}$
Hence, when the odds in favor of standing first of three students appearing in an examination are $1:2$, $2:5$ and $1:7$ respectively. The probability that either of them will stand first, is $\dfrac{{125}}{{168}}$.
Therefore, option A is the correct answer.

Note: Probability is the certainty of the occurrence of an event. A probability of an event can neither be less than 0 nor greater than 1. That means the probability of an event lies between 0 to 1. The probability of a sure event is always 1. Also, by mutually exclusive events, we mean those events which cannot occur at the same time i.e. all the three students cannot stand first in the examination at the same time.