Question

In a class there are x girls and y boys, a student is selected at random, then the probability of selecting a boy is
(a) $\dfrac{x}{y}$
(b) $\dfrac{x}{\left( x+y \right)}$
(c) $\dfrac{y}{\left( x+y \right)}$
(d) $\dfrac{y}{x}$

Answer 1

Hint: In this question, we first need to find the number of ways of selecting a student at random and the number of ways of selecting a boy using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$. Then, by using the formula $P=\dfrac{m}{n}$ divide the ways of selecting a boy with the total ways of selecting a student at random to get the probability.

Complete step-by-step answer:
PROBABILITY: If there are n elementary events associated with a random experiment and m of them are favourable to an event A, then the probability of happening or occurrence of A, denoted by P(A), is given by
$P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$
COMBINATION: Each of the different selections which can be made by all of a number of given things without reference to the order of the things is called a combination
The number of combinations is given by the formula
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
Now, there are x girls and y boys in the classroom
Now, the number of students in the classroom are $x+y$
Let us now select a student at random which is given by
$\Rightarrow {}^{\left( x+y \right)}{{C}_{1}}$
Now, this can be further written as
$\Rightarrow \dfrac{\left( x+y \right)!}{\left( x+y-1 \right)!1!}$
Now, on further simplification we get,
$\Rightarrow x+y$
Now, let us find the number of ways in which a boy can be selected
Here, there are y boys in the class out of which we need to select one
$\Rightarrow {}^{y}{{C}_{1}}$
Now, this can be further written as
$\Rightarrow \dfrac{y!}{\left( y-1 \right)!1!}$
Now, on further simplification we get,
$\Rightarrow y$
Let us now find the probability of selecting a student at random to be a boy
As we already know from the formula of probability that
$P\left( A \right)=\dfrac{m}{n}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$
Let us assume the event of selecting a student at random to be a boy
Here, m is the number of ways of selecting a boy and n is the number of ways of selecting a student a random
Now, we have from the values found above we have
$m=y,n=x+y$
Now, on substituting these values in the probability formula we get,
$\therefore P\left( A \right)=\dfrac{y}{x+y}$
Hence, the correct option is (c).

Note:Instead of finding the ways of selecting a student and a boy separately we can directly calculate them while calculating the probability at the same time. Both the methods give the same result but it would be a bit easy process.It is important to note that to find the favourable outcomes we need to find the number of ways of selecting a boy which is y but not x. Because there is a chance for choosing incorrect options as it is also mentioned.