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

Hint: 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
\[\begin{align}
  & \Rightarrow n={}^{z}{{C}_{1}} \\
 & \Rightarrow n=\dfrac{z!}{1!\left( z-1 \right)!} \\
 & \Rightarrow n=\dfrac{z\times \left( z-1 \right)!}{\left( z-1 \right)!}=z \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow k={}^{N}{{C}_{1}} \\
 & \Rightarrow k=\dfrac{N!}{1!\left( N-1 \right)!} \\
 & \Rightarrow k=\dfrac{N\times \left( N-1 \right)!}{\left( N-1 \right)!}=N \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow P=\dfrac{n}{k} \\
 & \Rightarrow P=\dfrac{z}{N} \\
\end{align}\]


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.