Question

A bag contains $x$ white, $y$ red and $z$ blue balls. A ball is drawn at random from the bag, then, the probability of getting a blue ball is:
(a) $\dfrac{x}{x+y+z}$
(b) $\dfrac{y}{x+y+z}$
(c) $\dfrac{z}{x+y+z}$
(d) 0

Answer 1

We solve this problem by using the simple formula of probability.
The formula of the probability is given as
$\Rightarrow P=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}$
For finding the number of possible number of outcomes we use the combinations that is the number of ways of selecting $'r'$ objects from $'n'$ objects is ${}^{n}{{C}_{r}}$ where
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Complete step by step answer:
We are given that there are $x$ white, $y$ red and $z$ blue balls in a bag.
Let us assume that the total number of balls in the bag as
$\Rightarrow N=x+y+z$
We are given that one ball is drawn at random from the bag.
We are asked to find the probability of getting a blue ball.
Let us assume that the number of possible outcomes of getting blue ball as $'n'$
Here, we can see that the number of possible outcomes of getting a blue ball is nothing but selecting 1 ball from $z$ blue balls.
We know that the number of ways of selecting $'r'$ objects from $'n'$ objects is ${}^{n}{{C}_{r}}$ where
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
By using the above formula we get

& \Rightarrow n={}^{z}{{C}_{1}} \\\ & \Rightarrow n=\dfrac{z!}{1!\left( z-1 \right)!} \\\ & \Rightarrow n=\dfrac{z\times \left( z-1 \right)!}{\left( z-1 \right)!}=z \\\ \end{aligned}$$ Now, let us calculate the total number of possible outcomes when a ball is drawn at random. We know that the total number of outcomes is nothing but selecting 1 ball from $$'N'$$ balls Let us assume that the total number of outcomes as $$k$$ then by using the combinations we get $$\begin{aligned} & \Rightarrow k={}^{N}{{C}_{1}} \\\ & \Rightarrow k=\dfrac{N!}{1!\left( N-1 \right)!} \\\ & \Rightarrow k=\dfrac{N\times \left( N-1 \right)!}{\left( N-1 \right)!}=N \\\ \end{aligned}$$ Let us assume that the probability of getting a blue ball as $$'P'$$ We know that the formula of the probability is given as $$\Rightarrow P=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}$$ By using the above formula we get $$\begin{aligned} & \Rightarrow P=\dfrac{n}{k} \\\ & \Rightarrow P=\dfrac{z}{N} \\\ \end{aligned}$$ Now, by substituting the value of $$'N'$$ we get $$\Rightarrow P=\dfrac{z}{x+y+z}$$ Therefore the required probability is $$\dfrac{z}{x+y+z}$$ **So, the correct answer is “Option c”.** **Note:** We can solve the problem directly without using the combinations. Here we are given that 1 ball is drawn at random. If there is only one ball drawn at random then the probability can be given as $$P=\dfrac{\text{number of required balls }}{\text{total number of balls}}$$ We are asked to find the probability of getting blue balls from $$x$$ white, $$y$$ red and $$z$$ blue balls. By using the above formula we get the required probability as $$\Rightarrow P=\dfrac{z}{x+y+z}$$ Therefore the required probability is $$\dfrac{z}{x+y+z}$$ So, option (c) is the correct answer.